Kindly Consider the following 103. 5x−21+2x+3x2

Let5x–2=Addx(1+2x+3x2)+B⇒ 5x–2=A(2+6x)+B=2A+A6x+B Equating thecoefficientof x andconstanttermonbothsides,wehave 5=6A; and 2A+B=–2 ⇒A= 56⇒ B=–2–2A B=−113 Therefore,5x−2=56(2+6x)+(−113)∴I=∫5x−21+2x+3x2dx=∫56(2+6x)+(−113)1+2x+3x2dx=56∫(2+6x)1+2x+3x2dx−113∫11+2x+3x2dxLet,I1=∫2+6x1+2x+3x2dx and I2=∫11+2x+3x2dx∴I=∫5x−21+2x+3x2dx=56 I1−113I2−−−−−(1)Let 1 + 2x+ 3x2=t ⇒(2+6x)dx=dtI1=∫dtt=log|t|=log|1+2x+3x2| ____(2)I2=∫11+2x+3x2dxNow,1+2x+3x2=1+3(x2+23x)Therefore,1+3(x2+23x)=1+3(x2+23x+19−19)

Kindly Consider the following 73. sin2x1+cosx

2x1+cosx= (2sinx2⋅cosx2)22cos2x2 =4sin2x2 cos2x22cos2x2=2sin2x22=1−cosx=∫sin2x1+cosx dx=∫ (1−cosx)dx=x−sinx+C

Kindly Consider the following 79. cos2x+2sin2xcos2x

=cos2x+ (1−cos2x)cos2x=1cos2x=sec2x=∫cos2x+2sin2xcos2xdx=∫sec2xd x = tanx+ C

Kindly Consider the following 24. (logx)2x

Let, logx=t1xdx=dtI=∫ (logx)2xdx=∫t2dt=t33+C= (logx)33+C

Kindly Consider the following 81. cos2x(cosx+sinx)2

Kindly Consider the following 83. 1cos(x−a)cos(x−b)

1cos(x−a)cos(x−b)=1sin(a−b)×[sin(a−b)cos(x−a)cos(x−b)]=1sin(a−b)[sin{(x−b)−(x−a)}cos(x−a)cos(x−b)] =1sin(a−b)[sin(x−b)cos(x−a)−cos(x−b)⋅sin(x−a)cos(x−a)cos(x−b)=1sin(a−b)[tan(x−b)−tan(x−a)]=1sin(a−b)∫ tan(x−b)−tan(x−a)]dx=1sin(a−b)[−log|cos(x−b)|+log|cos(x−a)?]=1sin(a−b)[log|cos(x−a)cos(x−b)|]+ C

Kindly Consider the following 85. ∫e2(1+x)cos2(e2x)

Lete2x=t e2x+ ex1dx=dt ex(x+1)dx=dt=∫ex(1+x)cos2(e2x)dx=∫dtcos2t=∫sec2t dt = tan (ex,x) + C ∴The correct answer is (B).

Kindly Consider the following 70. cosx1+cosx

cosx1+cosx=cos2x2−sin2x22cos2x2=12[1−tan2x2]=∫cosx1+cosx dx=12∫(1−tan2x2) dx=12(1−sec2x2+1)dx=12(2−sec2x2)dx=12[2x−tanx212]+ C=x−tanx2+ C

Kindly Consider the following 36. 1x(logx)m,x>0,m≠1

Put, logx=t⇒1xdx=dtI=∫1x (logx)mdx=∫dt (t)m= (t−m+11−m)+C= (logx)−m+1 (1−m)+C

Kindly Consider the following 74. cos2x−cos2∝cosx−cos∝

=4cos(x+∝2)cos(x−∝2)=2[cos(x+∝2+x−∝2)+cos(x−∝2−x−∝2)= 2[cos(x) + cos∝]=2cosx+2cos∝=∫cos2x−cos2∝cosx−cos∝ dx=∫(2cosx+2cos∝)dx = 2[sinx+xcos∝] + C

Kindly Consider the following 75. cosx−sinx1+sin2x

cosx−sinx1+sin2x=cosx−sinx(sin2x+cos2x)+2sinxcosx=cosx−sinx(sinx+cosx)2[?sin2+cos2=1 & sin2x=2sinxcosx]I=∫cosx−sinx1+sin2x dx=∫cosx−sinx(sinx+cosx)2dxPut sinx+ cosx=t (cosx−sinx)dx=dt=∫dtt2=∫t−2dt=−t−1+=−1t+ c=1−sinx+cosx+ c

Kindly Consider the following 41. e2x−1e2x+1

Dividing both numerator and denominator by ex, we get e2x−1exe2x+1ex=ex−e−xex+e−xPut ex+e−x=t⇒ (ex−e−x)dx=dtI=∫e2x−1e2x+1dx=∫ex−e−xex+e−xdxI=∫dtt=log|t|+c=log|ex+e−x|+c

Kindly Consider the following 42. e2x−e−2xe2x+e−2x

Put e2x+e−2x=t⇒ (2e2x−2e−2x)dx=dt ⇒2 (e2x−e−2x)dx=dtI=∫e2x−e−2xe2x+e−2xdx=∫dt2t=12∫1tdt=12log|t|+=12log|e2x+e−2x|+C

Kindly Consider the following 60. ∫10x9+10xloge10dxx10+10x

Put x10+10x=t⇒ (10x9+10xloge10)dx=dtI=∫10x9+10xloge10x10+10xdx=∫dtt=logt+C=log (10x+x10)+c Therefore, the correct answer is (D)

Kindly Consider the following 61. ∫dxsin2xcos2xdx

I=∫dxsin2xcos2x=∫1sin2xcos2xdx=∫sin2x+cos2xsin2xcos2xdx=∫sin2xsin2xcos2xdx+∫cos2xsin2xcos2xdx=∫sec2x dx+∫cosec2x−dx]=tanx−cotx+c Therefore, the correct answer is B.

Kindly Consider the following 7. ∫x2(1−1x2)dx

∫x2−x2x2dx=∫x2–1dx=∫x2dx–∫1dx =x33−x+C

Kindly Consider the following 8. ∫(ax2+bx+c)dx

=a.∫x2.dx+b∫x.dx+c∫1.dx=a·x33+bx22+cx+d.

Kindly Consider the following 9. ∫(2x2+ex)dx

∫2x2+exdx =2·∫x2dx+∫exdx=2·x33+ex+ C

Kindly Consider the following 11. ∫x3+5x2−4x2dx

∫ (x3x2+5x2x2−4x2)dx=∫ (x+5−4x–2)dx =∫x.dx+∫5dx–∫4x–2dx=x22+5x−4x−2+1−2+1=x22+5x−4x−1−1=x22+5x+4x+ C

Kindly Consider the following 13. ∫x3−x2+x−1x−1dx

∫x3−x2x−1+x−1x−1dx∫ (x2 (x−1)x−1+1)dx=∫ (x2+1)dx =∫x2.dx+∫1.dx=x33+x+ C

Kindly Consider the following 14. ∫(1−x)√xdx

∫ (√x−x·x1/2)dx∫ (√x−x3/2)·dx∫x1/2·dx−∫x3/2·dxx3/23/2−x5/25/2+C23x3/2−25x5/2+ c

Kindly Consider the following 15. ∫√x(3x2+2x+3)dx

=∫(3x2.x1/2+2x.x1/2+3x1/2).dz =∫(3x5/2+2x3/2+3x4/2).dx =∫3x5/2.dx+∫2x3/2.dx+∫3x1/2.dx=3·x7/27/2+2·x5/25/2+3·x3/23/2+ C=6·x7/27+4x5/25+63x3/2+ C=6x7/27+4·x5/25+2x3/2+ C.

Kindly Consider the following 16. ∫(2x–3cosx+ex)dx

=∫2xdx–∫3cosx+exdx =2x22−3sinx+ex+ C=x2−3sinx+ex+C

Kindly Consider the following 17. ∫(2x2–3sinx+5√x)dx

=∫2x2.dx–∫3sinx.dx+∫5√xdx=2x33+3cosx+5x3/232+ C=2x33+3cosx+10x3/23+ C

Kindly Consider the following 18. ∫secx(secx+tanx)dx

=∫ (sec2x+secxtanx)dx =∫sec2x.dx+∫secxtanxdx =tanx+secx+C

Kindly Consider the following 19. ∫sec2xcosec2xdx

∫1cos2x1cos2xdx=∫sin2xcos2xdx=∫tan2xdx =∫ (sec2x–1)dx=∫sec2xdx–∫1.dx =tanx–x+C

Kindly Consider the following 20. ∫2−3sinxcos2xdx

∫ (2cos2x−3sinxcos2x)dx=∫ (2sec2x–3tanx.secx)dx =∫2sec2x.dx–∫3tanx.secx.dx =2tanx– 3 secx+C

Kindly Consider the following 22. ddxf(x)=4x3−3x4,f(2)=0

f (x)=∫4x3–3x4dx=4⋅x44−3⋅x−3−3+ C=x4+1x3+ C Now, f (2)=0 ∴f (2) =24+123+C=0 =16+18+c=0=c=− (16+18)=c=−1298 Therefore, correct answer is A.

Class 12 Maths Chapter 7 NCERT Solutions: Download Integrals Key Topics & Formulas

Q: Kindly Consider the following 12.

=∫(x3.x–1/2+3x.x–1/2+4.x–1/2)dx=∫(x5/2+3x1/2+4x−1/2)dx=x52+152+1+3·x1/2+112+1+4·x−1/2+1−12+1=x7/27/2+3·x3/23/2+4x1/21/2+C=27x7/2+2x3/2+8x1/2+C=27x7/2+2x3/2+8√x+C.

Q: Kindly Consider the following 27. sin(ax+b)cos(ax+b)

=sin (ax+b)cos (ax+b)=2sin (ax+b)cos (ax+b)2=sin2 (ax+b)2Put 2 (ax+b)=t2adx=dtI=∫sin2 (ax+b)2dx=12∫sint dt2a=14a [−cost]+C=−14acos2 (ax+b)+C

Q: Kindly Consider the following 32.

x−√x=1√x (√x−1)Put, √x−1=t12√xdx=dtI=∫1√x (√x−1)dx=∫2t dt=2log|t|+c=2log (√x−1)+c

Q: Kindly Consider the following 77. tan4x

tan4x=tan2xtan2x =(sec2x−1)tan2x =sec2x.tan2x−tan2x =sec2x.tan2x−(sec2x−1) I =sec2x.tan2x−sec2x+1 I=∫tan4xdx=∫sec2x.tan2xdx−∫sec2xdx+∫1.dx =∫sec2x.tan2xdx−tanπ+x+C−−−−−(i) LetI=∫sec2x.tan2xdx Put tanx=t sec2xdx=dt I1=∫sec2x.tan2xdx =∫t2dt =t33=tan3x 3I=∫tan4xdx =13tan3x−tanx+x+ C

Q: Kindly Consider the following 82. sin−1(cosx)

I=∫sin−1 (cosx)dx=∫sin−1 (sin {π2−x})dx=∫ {π2−x}dx=π2x−x22+ C

Updated on Aug 5, 2025 12:29 IST

By Pallavi Pathak, Assistant Manager Content

Integrals Class 12 NCERT Solutions cover two types of integrals - definite and indefinite integrals, which are together called the Integral Calculus. Between the indefinite and definite integrals, there is a connection called the Fundamental Theorem of Calculus. This connection makes the definite integral a practical tool for engineering and science.
Class 12 Integrals concepts are important as it has applications in various industries. The definite integral is used to solve many interesting problems from various fields like probability, finance, and economics. The chapter also includes the elementary properties of the definite and indefinite integrals, including some techniques of integration.
If you are looking for Class 12 Maths notes for CBSE Board exam preparation, check - Class 12 Maths Notes. You will get the solved examples with chapter-wise PDFs.

Table of contents

Insight into Class 12 Maths Chapter 7 Integrals NCERT Solutions
Class 12 Math Chapter 7 Integral: Key Topics, Weightage
Important Formulas of Class 12 Integrals
Topics Covered in NCERT Maths Class 12 Integrals Chapter
NCERT Maths Class 12th Solution PDF - Integrals Chapter Download
Integrals Questions and Answers

Insight into Class 12 Maths Chapter 7 Integrals NCERT Solutions

Here is a quick review of the Integrals Class 12:

The chapter is about integration, which is the inverse process of differentiation. Here, the differential of a function is given, and we need to find the function. $\frac{d}{d x} F (x) = f (x) Then we write \int_{f (x)}^{F (x) + C} d x$
Some properties of indefinite integrals are - $\int [f (x) + g (x)] d x = \int f (x) d x + \int g (x) d x$ $k \int f (x) d x = \int k f (x) d x More generally, \int [k_{1} f_{1} (x) + k_{2} f_{2} (x) + \dots + k_{n} f_{n} (x)] d x = k_{1} \int f_{1} (x) d x + k_{2} \int f_{2} (x) d x + \dots + k_{n} \int f_{n} (x) d x$
The chapter also includes the standard integrals, integration by partial function, integration by substitution, and integrals of some special functions.
Other concepts covered in this chapter include integration by parts, some special types of integrals, and the first fundamental theorem of integral calculus.

Class 12 Math Chapter 7 Integral: Key Topics, Weightage

Class 12 Integrals is an important chapter for entrance tests. Students should clearly understand the concepts of the chapter to score well in the CBSE Board exam and competitive exams like JEE Mains. Here are the topics covered in this chapter:

Exercise	Topics Covered
7.1	Introduction
7.2	Integration as an Inverse Process of Differentiation
7.3	Methods of Integration
7.4	Integrals of Some Particular Functions
7.5	Integration by Partial Fractions
7.6	Integration by Parts
7.7	Definite Integral
7.8	Fundamental Theorem of Calculus
7.9	Evaluation of Definite Integrals by Substitution
7.10	Some Properties of Definite Integrals

Class 12 Integrals Weightage in JEE Main

Exam	Number of Questions	Total Marks	Weightage
JEE Main	3 questions	Generally worth around 12 marks	9-10%

Those who are looking for comprehensive NCERT notes of Physics, Chemistry & Maths of class 12, must explore at - NCERT Class 12 Notes.

Important Formulas of Class 12 Integrals

Important Formulae for Class 12 Math Integrals

$\int e^x \, dx = e^x + C$
$\int a^x \, dx = \frac{a^x}{\ln a} + C$
$\int \frac{1}{x} \, dx = \ln |x| + C$
$\int \sin x \, dx = -\cos x + C$
$\int \cos x \, dx = \sin x + C$
$\int \sec^2 x \, dx = \tan x + C$
$\int \csc^2 x \, dx = -\cot x + C$
$\int \sec x \tan x \, dx = \sec x + C$
$\int \csc x \cot x \, dx = -\csc x + C$

Topics Covered in NCERT Maths Class 12 Integrals Chapter

Integration as an Inverse Process of Differentiation
Methods of Integration
Integrals of Some Particular Functions
Integration by Partial Fractions
Integration by Parts
Definite Integral
Fundamental Theorem of Calculus
Evaluation of Definite Integrals by Substitution
Some Properties of Definite Integrals

NCERT Maths Class 12th Solution PDF - Integrals Chapter Download

Integrals Class 12 NCERT Solutions PDF download link is given here. Students should download it to get the well-structured solutions to all the exercises of this chapter. It offers reliable and accurate study material for exam preparation.

Download Here: NCERT Solution for Class XII Maths Integrals PDF

Integrals Questions and Answers

Find an anti-derivative (or integral) of the following functions by the method of inspection.

Q1. $s i n 2 x$

A.1.

$\begin{array}{l} \frac{d}{d x} c o s 2 x = - 2 s i n 2 x \\ s i n 2 x = - \frac{1}{2} \frac{d}{d x} (c o s 2 x) \\ s i n 2 x = \frac{d}{d x} (\frac{1}{2} c o s 2 x) \end{array}$

Therefore, an anti-derivative of $s i n 2 x i s - \frac{1}{2} c o s 2 x$

Q2. $c o s 3 x$

A.2.

$\begin{array}{l} \frac{d}{d x} s i n 3 x = 3 c o s 3 x \\ c o s 3 x = \frac{1}{3} d x (s i n 3 x) \\ c o s 3 x = \frac{d}{d x} (\frac{1}{3} s i n 3 x) \end{array}$

Therefore, an anti-derivative of $c o s 3 x i s \frac{1}{3} s i n 3 x$

Q3. $e^{2 x}$

A.3.

$\begin{array}{l} \frac{d}{d x} (e^{2} x) = 2 e^{2 x} \\ e^{2 x} = \frac{1}{2} - \frac{d}{d x} (e^{2 x}) \\ e^{2 x} = \frac{d}{d x} (\frac{1}{2} e^{2 x}) \end{array}$

Therefore, an anti-derivative of $e^{2 x} i s \frac{1}{2} e^{2 x}$

Q4. ${(a x + b)}^{2}$

A.4.

$\frac{d}{d x} {(a x + b)}^{3} = 3 a {(a x + b)}^{2}$

${(a x + b)}^{2} = \frac{1}{3 a} . \frac{a}{d x} {(a x + b)}^{3}$

${(a x + b)}^{2} = \frac{d}{d x} (\frac{1}{3 a} {(a x + b)}^{3})$

Therefore, an anti-derivative of ${(a x + b)}^{3} i s \frac{1}{3 a} (a x + b)^{3}$

Kindly Consider the following

103. $\frac{5 x - 2}{1 + 2 x + 3 x^{2}}$

$\begin{array}{l} L e t 5 x - 2 = A \frac{d}{d x} (1 + 2 x + 3 x^{2}) + B \\ \Rightarrow 5 x - 2 = A (2 + 6 x) + B \\ = 2 A + A 6 x + B \end{array}$

$\begin{array}{l} E q u a t i n g t h e c o e f f i c i e n t o f x a n d c o n s t a n t t e r m o n b o t h s i d e s, w e h a v e \\ 5 = 6 A; a n d 2 A + B = - 2 \\ \Rightarrow A = \frac{5}{6} \\ \Rightarrow B = - 2 - 2 A \\ B = \frac{- 11}{3} \\ T h e r e f o r e, \\ 5 x - 2 = \frac{5}{6} (2 + 6 x) + (\frac{- 11}{3}) \\ ∴ I = \int \frac{5 x - 2}{1 + 2 x + 3 x^{2}} d x \\ = \int \frac{\frac{5}{6} (2 + 6 x) + (\frac{- 11}{3})}{1 + 2 x + 3 x^{2}} d x \\ = \frac{5}{6} \int \frac{(2 + 6 x)}{1 + 2 x + 3 x^{2}} d x - \frac{11}{3} \int \frac{1}{1 + 2 x + 3 x^{2}} d x \\ L e t, I_{1} = \int \frac{2 + 6 x}{1 + 2 x + 3 x^{2}} d x a n d I_{2} = \int \frac{1}{1 + 2 x + 3 x^{2}} d x \\ ∴ I = \int \frac{5 x - 2}{1 + 2 x + 3 x^{2}} d x \\ = \frac{5}{6} I_{1} - \frac{11}{3} I_{2} - - - - - (1) \\ \begin{array}{l} L e t 1 + 2 x + 3 x^{2} = t \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} \Rightarrow (2 + 6 x) d x = d t \end{array} \\ I_{1} = \int \frac{d t}{t} = l o g | t | \\ = l o g | 1 + 2 x + 3 x^{2} |____(2) \\ I_{2} = \int \frac{1}{1 + 2 x + 3 x^{2}} d x \\ N o w, 1 + 2 x + 3 x^{2} = 1 + 3 (x^{2} + \frac{2}{3} x) \\ T h e r e f o r e, 1 + 3 (x^{2} + \frac{2}{3} x) = 1 + 3 (x^{2} + \frac{2}{3} x + \frac{1}{9} - \frac{1}{9}) \end{array}$

Kindly Consider the following

73. $\frac{s i n^{2} x}{1 + c o s x}$

$\frac{2 x}{1 + c o s x} = \frac{{(2 s i n \frac{x}{2} \cdot c o s \frac{x}{2})}^{2}}{2 c o s^{2} \frac{x}{2}}$

$\begin{array}{l} = \frac{4 s i n^{2} \frac{x}{2} c o s^{2} \frac{x}{2}}{2 c o s^{2} \frac{x}{2}} = \frac{2 s i n^{2} \frac{x}{2}}{2} \\ = 1 - c o s x \\ = \int \frac{s i n^{2} x}{1 + c o s x} d x \\ = \int (1 - c o s x) d x \\ = x - s i n x + C \end{array}$

Kindly Consider the following

79. $\frac{c o s 2 x + 2 s i n^{2} x}{c o s^{2} x}$

$\begin{array}{l} = \frac{c o s 2 x + (1 - c o s 2 x)}{c o s^{2} x} \\ = \frac{1}{c o s^{2} x} = s e c^{2} x \\ = \int \frac{c o s 2 x + 2 s i n^{2} x}{c o s^{2} x} d x \\ \begin{array}{l} = \int s e c^{2} x d x \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} = t a n x + C \end{array} \end{array}$

Kindly Consider the following

$\begin{array}{l} = \int (x^{3} . x^{- 1 / 2} + 3 x . x^{- 1 / 2} + 4 . x^{- 1 / 2}) d x \\ = \int (x^{5 / 2} + 3 x^{1 / 2} + 4 x^{- 1 / 2}) d x \\ = \frac{x^{\frac{5}{2} + 1}}{\frac{5}{2} + 1} + 3 \cdot \frac{x^{1 / 2 + 1}}{\frac{1}{2} + 1} + 4 \cdot \frac{x^{- 1 / 2 + 1}}{- \frac{1}{2} + 1} \\ = \frac{x^{7 / 2}}{7 / 2} + \frac{3 \cdot x^{3 / 2}}{3 / 2} + \frac{4 x^{1 / 2}}{1 / 2} + C \\ = \frac{2}{7} x^{7 / 2} + 2 x^{3 / 2} + 8 x^{1 / 2} + C \\ = \frac{2}{7} x^{7 / 2} + 2 x^{3 / 2} + 8√x + C . \end{array}$

Kindly Consider the following

24. $\frac{{(l o g x)}^{2}}{x}$

$\begin{array}{l} L e t, l o g x = t \\ \frac{1}{x} d x = d t \\ I = \int \frac{{(l o g x)}^{2}}{x} d x = \int t^{2} d t = \frac{t^{3}}{3} + C \\ = \frac{{(l o g x)}^{3}}{3} + C \end{array}$

Kindly Consider the following

27. $s i n (a x + b) c o s (a x + b)$

$\begin{array}{l} = s i n (a x + b) c o s (a x + b) = \frac{2 s i n (a x + b) c o s (a x + b)}{2} \\ = \frac{s i n 2 (a x + b)}{2} \\ P u t 2 (a x + b) = t \\ 2 a d x = d t \\ I = \int \frac{s i n 2 (a x + b)}{2} d x \\ = \frac{1}{2} \int \frac{s i n t d t}{2 a} = \frac{1}{4 a} [- c o s t] + C \\ = - \frac{1}{4 a} c o s 2 (a x + b) + C \end{array}$

Kindly Consider the following

32.

$\begin{array}{l} x -\sqrtx = \frac{1}{\sqrtx (\sqrtx - 1)} \\ P u t, \sqrtx - 1 = t \\ \frac{1}{2√x} d x = d t \\ I = \int \frac{1}{\sqrtx (\sqrtx - 1)} d x = \int \frac{2}{t} d t \\ = 2 l o g | t | + c \\ = 2 l o g (\sqrtx - 1) + c \end{array}$

Kindly Consider the following

77. $t a n^{4} x$

$t a n^{4} x = t a n^{2} x t a n^{2} x$

$= (s e c^{2} x - 1) t a n^{2} x$

$= s e c^{2} x . t a n^{2} x - t a n^{2} x$

$= s e c^{2} x . t a n^{2} x - (s e c^{2} x - 1)$

I $\begin{array}{l} \begin{array}{l} \end{array} = s e c^{2} x . t a n^{2} x - s e c^{2} x + 1 \end{array}$

$\begin{array}{l} I = \int t a n^{4} x d x = \int s e c^{2} x . t a n^{2} x d x - \int s e c^{2} x d x + \int 1 . d x \end{array}$

$\begin{array}{l} = \int s e c^{2} x . t a n^{2} x d x - t a n π + x + C - - - - - (i) \end{array}$

$\begin{array}{l} L e t I = \int s e c^{2} x . t a n^{2} x d x \end{array}$

$\begin{array}{l} P u t t a n x = t \end{array}$

$\begin{array}{l} s e c^{2} x d x = d t \end{array}$

$\begin{array}{l} I_{1} = \int s e c^{2} x . t a n^{2} x d x \end{array}$

$\begin{array}{l} = \int t^{2} d t \end{array}$

$= \frac{t^{3}}{3} = t a n^{3} x$

$3 I = \int t a n^{4} x d x$

$= \frac{1}{3} t a n^{3} x - t a n x + x + C$

Kindly Consider the following

81. $\frac{c o s 2 x}{{(c o s x + s i n x)}^{2}}$

$\begin{array}{l} = \frac{c o s 2 x}{c o s^{2} x + s i n^{2} x + 2 s i n x c o s x} \\ = \frac{c o s 2 x}{1 + s i n 2 x} \end{array}$

$\begin{array}{l} = \int \frac{c o s 2 x}{{(c o s x + s i n x)}^{2}} d x \\ = \int \frac{c o s 2 x}{1 + s i n 2 x} d x \\ \begin{array}{l} P u t 1 + s i n 2 x = t \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} 2 c o s 2 x d x = d t \end{array} \\ = \int \frac{c o s 2 x}{{(c o s x + s i n x)}^{2}} d x \\ = \frac{1}{2} \int \frac{1}{t} d t = \frac{1}{2} l o g | t | + C \\ = \frac{1}{2} l o g | 1 + s i n 2 x | + C \\ - \frac{1}{2} l o g | {(c o s x + s i n x)}^{2} | + C \\ = l o g | c o s x + s i n x | + C \end{array}$

Kindly Consider the following

82. $s i n^{- 1} (c o s x)$

$\begin{array}{l} I = \int s i {n^{-}}^{1} (c o s x) d x \\ = \int s i n^{- 1} (s i n {\frac{π}{2} - x}) d x \\ = \int {\frac{π}{2} - x} d x \\ = \frac{π}{2} x - \frac{x^{2}}{2} + C \end{array}$

Kindly Consider the following

83. $\frac{1}{c o s (x - a) c o s (x - b)}$

$\begin{array}{l} \frac{1}{c o s (x - a) c o s (x - b)} = \frac{1}{s i n (a - b)} \times [\frac{s i n (a - b)}{c o s (x - a) c o s (x - b)}] \\ = \frac{1}{s i n (a - b)} [\frac{s i n {(x - b) - (x - a)}}{c o s (x - a) c o s (x - b)}] \end{array}$

$\begin{array}{l} = \frac{1}{s i n (a - b)} [s i n (x - b) c o s (x - a) - c o s (x - b) \cdot s i n (x - a) c o s (x - a) c o s (x - b) \\ = \frac{1}{s i n (a - b)} [t a n (x - b) - t a n (x - a)] \\ = \frac{1}{s i n (a - b)} \int t a n (x - b) - t a n (x - a)] d x \\ = \frac{1}{s i n (a - b)} [- l o g | c o s (x - b) | + l o g | c o s (x - a) ?] \\ = \frac{1}{s i n (a - b)} [l o g | \frac{c o s (x - a)}{c o s (x - b)} |] + C \end{array}$

Kindly Consider the following

85. $\int \frac{e^{2} (1 + x)}{c o s^{2} (e^{2} x)}$

$\begin{array}{l} \begin{array}{l} L e t e^{2} x = t \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} e^{2} x + e^{x} 1 d x = d t \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} e^{x} (x + 1) d x = d t \end{array} \\ = \int \frac{e^{x} (1 + x)}{c o s^{2} (e^{2} x)} d x = \int \frac{d t}{c o s^{2} t} \\ \begin{array}{l} = \int s e c^{2} t d t = t a n (e^{x}, x) + C \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} ∴ T h e c o r r e c t a n s w e r i s (B) . \end{array} \end{array}$

87. Kindly Consider the following

Kindly go through the solution

88. Kindly Consider the following

Kindly go through the solution

Kindly Consider the following

144.

Kindly go through the solution

Kindly Consider the following

145. $x s e c^{2} x$

$\begin{array}{l} \int x s e c^{2} x d x = x \int s e c^{2} x d x - \int \frac{d x}{d x} \int s e c^{2} x d x d x . \\ = x t a n x - \int t a n x d x \\ = x t a n x - (- l o g | c o s x |) + C \\ = x t a n x + l o g | c o s x | + C \end{array}$

251. Kindly Consider the following

Kindly go through the solution

Kindly Consider the following

70. $\frac{c o s x}{1 + c o s x}$

$\begin{array}{l} \frac{c o s x}{1 + c o s x} = \frac{c o s^{2} \frac{x}{2} - s i n^{2} \frac{x}{2}}{2 c o s^{2} \frac{x}{2}} \\ = \frac{1}{2} [1 - t a n^{2} \frac{x}{2}] \\ = \int \frac{c o s x}{1 + c o s x} d x = \frac{1}{2} \int (1 - t a n^{2} \frac{x}{2}) d x \\ = \frac{1}{2} (1 - s e c^{2} \frac{x}{2} + 1) d x \\ = \frac{1}{2} (2 - s e c^{2} \frac{x}{2}) d x = \frac{1}{2} [2 x - \frac{t a n \frac{x}{2}}{\frac{1}{2}}] + C \\ = x - t a n \frac{x}{2} + C \end{array}$

ind an anti-derivative (or integral) of the following functions by the method of inspection.

1. $s i n 2 x$

$\begin{array}{l} \frac{d}{d x} c o s 2 x = - 2 s i n 2 x \\ s i n 2 x = - \frac{1}{2} \frac{d}{d x} (c o s 2 x) \\ s i n 2 x = \frac{d}{d x} (\frac{1}{2} c o s 2 x) \end{array}$

Therefore, an anti-derivative of $s i n 2 x i s - \frac{1}{2} c o s 2 x$

Kindly Consider the following

28. √ax + b

$\begin{array}{l} P u t, a x + b = t \\ a d x = d t \\ d x = \frac{1}{a} d t \\ I = \int (a x + b)^{\frac{1}{2}} d x \\ = \int t^{\frac{1}{2}} \cdot \frac{1}{a} d t = \frac{1}{a} \frac{t \frac{1}{2 + 1}}{\frac{1}{2 + 1}} + C \\ = \frac{1}{a} \frac{t^{\frac{3}{2}}}{\frac{3}{2}} + C \\ = \frac{2}{3 a} {(a x + b)}^{\frac{3}{2}} + c \end{array}$

Kindly Consider the following

29. $x√x + 2$

$\begin{array}{l} P u t, x + 2 = t \\ d x = d t \\ \Rightarrow x = t - 2 \\ I = \int x√x+2 d x \\ = \int (t - 2) √t d t \\ = \int (t^{1 + \frac{3}{2}} - 2 t^{\frac{1}{2}}) d t \\ = \int (t^{\frac{3}{2}} - 2 t^{\frac{1}{2}}) d t \\ = \int t^{\frac{3}{2}} \cdot d t - \int 2 \cdot t^{\frac{1}{2}} \cdot d t \\ = \frac{t^{\frac{5}{2}}}{\frac{5}{2}} - 2 \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \\ = \frac{2}{5} t^{\frac{5}{2}} - \frac{4}{3} t^{\frac{3}{2}} + C \\ = \frac{2}{5} {(x + 2)}^{\frac{5}{2}} - \frac{4}{3} {(x + 2)}^{\frac{3}{2}} + C \end{array}$

Kindly Consider the following

36. $\frac{1}{x {(l o g x)}^{m}}, x > 0, m \neq 1$

$\begin{array}{l} P u t, l o g x = t \Rightarrow \frac{1}{x} d x = d t \\ I = \int \frac{1}{x {(l o g x)}^{m}} d x = \int \frac{d t}{{(t)}^{m}} \\ = (\frac{t^{- m + 1}}{1 - m}) + C = \frac{{(l o g x)}^{- m + 1}}{(1 - m)} + C \end{array}$

Kindly Consider the following

71. $s i n^{4} x$

$\begin{array}{l} s i n^{4} x = s i n^{2} x . s i n^{2} x \\ = (\frac{1 - c o s 2 x}{2}) (\frac{1 - c o s 2 x}{2}) \\ = \frac{1}{4} {(1 - c o s 2 x)}^{2} = \frac{1}{4} [1 + c o s^{2} 2 x - 2 c o s 2 x] \\ = \frac{1}{4} [1 + (\frac{1 + c o s 4 x}{2}) - 2 c o s 2 x] \\ = \frac{1}{4} [1 + \frac{1}{2} + \frac{1}{2} c o s 4 x - 2 c o s 2 x] \\ = \frac{1}{4} [\frac{3}{2} + \frac{1}{2} c o s 4 x - 2 c o s 2 x] \end{array}$

$\begin{array}{l} = \int s i n^{4} x d x = \frac{1}{4} \int [\frac{3}{2} + \frac{1}{2} c o s 4 x - 2 c o s 2 x] d x \\ = \frac{1}{4} [\frac{3}{2} + \frac{1}{2} (\frac{s i n 4 x}{4}) - \frac{2 s i n 2 x}{2}] + C \\ = \frac{1}{8} [3 x + \frac{s i n^{4} x}{4} - 2 s i n 2 x] + C \\ = \frac{3 x}{8} - \frac{1}{4} s i n 2 x + \frac{1}{32} s i n 4 x + C \end{array}$

Kindly Consider the following

72. $c o s^{4} 2 x$

Kindly go through the solution

Kindly Consider the following

74. $\frac{c o s 2 x - c o s 2 \propto}{c o s x - c o s \propto}$

$\begin{array}{l} = 4 c o s (\frac{x + \propto}{2}) c o s (\frac{x - \propto}{2}) \\ = 2 [c o s (\frac{x + \propto}{2} + \frac{x - \propto}{2}) + c o s (\frac{x - \propto}{2} - \frac{x - \propto}{2}) \\ \begin{array}{l} = 2 [c o s (x) + c o s \propto] & \begin{array}{l} = 2 c o s x + 2 c o s \propto \\ = \int \frac{c o s 2 x - c o s 2 \propto}{c o s x - c o s \propto} d x \\ \begin{array}{l} = \int (2 c o s x + 2 c o s \propto) d x \end{array} \end{array} \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} \begin{array}{l} \begin{array}{l} = 2 [s i n x + x c o s \propto] + C \end{array} \end{array} \end{array} \end{array}$

Kindly Consider the following

75. $\frac{c o s x - s i n x}{1 + s i n 2 x}$

$\begin{array}{l} \frac{c o s x - s i n x}{1 + s i n 2 x} = \frac{c o s x - s i n x}{(s i n^{2} x + c o s^{2} x) + 2 s i n x c o s x} \\ = \frac{c o s x - s i n x}{{(s i n x + c o s x)}^{2}} \\ [? s i n^{2} + c o s^{2} = 1 & s i n 2 x = 2 s i n x c o s x] \\ I = \int \frac{c o s x - s i n x}{1 + s i n 2 x} d x = \int \frac{c o s x - s i n x}{{(s i n x + c o s x)}^{2}} d x \\ \begin{array}{l} P u t s i n x + c o s x = t \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} (c o s x - s i n x) d x = d t \end{array} \\ = \int \frac{d t}{t^{2}} = \int t^{- 2} d t \\ = - t^{- 1} + = - \frac{1}{t} + c \\ = \frac{1}{- s i n x + c o s x} + c \end{array}$

Kindly Consider the following

76. $t a n^{3} 2 x . s e c 2 x$

$\begin{array}{l} \begin{array}{l} t a n^{3} 2 x s e c 2 x = t a n^{2} 2 x t a n 2 x s e c 2 x \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} {s e c^{2} (2 x) - 1} t a n 2 x s e c 2 x \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} = s e c^{2} 2 x t a n 2 x s e c 2 x \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} - t a n 2 x s e c 2 x \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} I = \int t a n^{3} 2 x . s e c 2 x d x \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} = \int t a n^{2} 2 x t a n 2 x s e c 2 x d x - \int t a n 2 x . s e c 2 x d x \end{array} \\ = \int t a n^{2} 2 x t a n 2 x s e c 2 x d x - \frac{s e c 2 x}{2} + \\ \begin{array}{l} P u t s e c 2 x = t \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} \Rightarrow 2 s e c 2 x t a n 2 x d x = d t \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} I = {\int t a n}^{3} 2 x . s e c 2 x d x \end{array} \\ = \frac{1}{2} \int t^{2} d t - \frac{s e c 2 x}{2} + C \\ = \frac{t^{3}}{6} - \frac{s e c 2 x}{2} + C \\ = \frac{{(s e c 2 x)}^{3}}{6} - \frac{s e c 2 x}{2} + C \end{array}$

Find the following integrals in Exercises 6 to 20:

6. $\int (4 e^{3 x} + 1) d x$

= $\begin{array}{l} \int 4 e^{3 x} d x + \int 1 d x \\ \frac{4 \cdot e^{3 x}}{3} + x + C \end{array}$

Kindly Consider the following

41. $\frac{e^{2 x} - 1}{e^{2 x} + 1}$

Dividing both numerator and denominator by e^x, we get

$\begin{array}{l} \frac{\frac{e^{2 x} - 1}{e^{x}}}{\frac{e^{2 x} + 1}{e^{x}}} = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} \\ P u t e^{x} + e^{- x} = t \\ \Rightarrow (e^{x} - e^{- x}) d x = d t \\ I = \int \frac{e^{2 x} - 1}{e^{2 x} + 1} d x = \int \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} d x \\ I = \int \frac{d t}{t} = l o g | t | + c \\ = l o g | e^{x} + e^{- x} | + c \end{array}$

Kindly Consider the following

42. $\frac{e^{2 x} - e^{- 2 x}}{e^{2 x} + e^{- 2 x}}$

$\begin{array}{l} P u t e^{2 x} + e^{- 2 x} = t \\ \begin{array}{l} \Rightarrow (2 e^{2 x} - 2 {e^{-}}^{2 x}) d x = d t \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} \Rightarrow 2 (e^{2 x} - {e^{-}}^{2 x}) d x = d t \end{array} \\ I = \int \frac{e^{2 x} - e^{- 2 x}}{e^{2 x} + e^{- 2 x}} d x = \int \frac{d t}{2 t} = \frac{1}{2} \int \frac{1}{t} d t \\ = \frac{1}{2} l o g | t | + = \frac{1}{2} l o g | e^{2 x} + e^{- 2 x} | +C \end{array}$

Kindly Consider the following

60. $\int \frac{10 x^{9} + 1 0^{x} l o g_{e} 10 d x}{x^{10} + 1 0^{x}}$

$\begin{array}{l} P u t x^{10} + 1 0^{x} = t \\ \Rightarrow (10 x^{9} + 1 0^{x} l o g^{e} 10) d x = d t \\ I = \int \frac{10 x^{9} + 1 0^{x} l o g_{e} 10}{x^{10} + 1 0^{x}} d x \\ = \int \frac{d t}{t} = l o g t + C \\ = l o g (1 0^{x} + x^{10}) + c \end{array}$

Therefore, the correct answer is (D)

Kindly Consider the following

61. $\int \frac{d x}{s i n^{2} x c o s^{2} x} d x$

$\begin{array}{l} I = \int \frac{d x}{s i n^{2} x c o s^{2} x} \\ = \int \frac{1}{s i n^{2} x c o s^{2} x} d x \\ = \int \frac{s i n^{2} x + c o s^{2} x}{s i n^{2} x c o s^{2} x} d x \\ = \int \frac{s i n^{2} x}{s i n^{2} x c o s^{2} x} d x + \int \frac{c o s^{2} x}{s i n^{2} x c o s^{2} x} d x \\ = \int s e c^{2} x d x + \int c o s e c^{2} x - d x] \\ = t a n x - c o t x + c \end{array}$

Therefore, the correct answer is B.

Kindly Consider the following

2. $c o s 3 x$

$\begin{array}{l} d x s i n 3 x = 3 c o s 3 x \\ c o s 3 x = \frac{1}{3} d x (s i n 3 x) \\ c o s 3 x = \frac{d}{d x} (\frac{1}{3} s i n 3 x) \end{array}$

Therefore, an anti-derivative of $c o s 3 x i s \frac{1}{3} s i n 3 x$

Kindly Consider the following

3. $e^{2 x}$

$\begin{array}{l} \frac{d}{d x} (e^{2} x) = 2 e^{2 x} \\ e^{2 x} = \frac{1}{2} - \frac{d}{d x} (e^{2 x}) \\ e^{2 x} = \frac{d}{d x} (\frac{1}{2} e^{2 x}) \end{array}$

Therefore, an anti-derivative of $e^{2 x} i s \frac{1}{2} e^{2 x}$

Kindly Consider the following

4. ${(a x + b)}^{2}$

$d x {(a x + b)}^{3} = 3 a {(a x + b)}^{2}$

${(a x + b)}^{2} = \frac{1}{3 a} . \frac{a}{d x} {(a x + b)}^{3}$

${(a x + b)}^{2} = \frac{d}{d x} (\frac{1}{3 a} {(a x + b)}^{3})$

Therefore, an anti-derivative of ${(a x + b)}^{3} i s \frac{1}{3 a} (a x + b)^{3}$

Kindly Consider the following

5. $s i n 2 x - 4 e 3 x$

$\frac{d}{d x} (- \frac{1}{2} c o s 2 x - \frac{4}{3} e^{3 x}) = s i n 2 x - 4 e^{3 x}$

Therefore, an anti-derivative of $(s i n 2 x - 4 e^{3 x}) i s - \frac{1}{2} c o s 2 x - \frac{4}{3} e^{3 x}$

Kindly Consider the following

7. $\int x^{2} (1 - \frac{1}{x^{2}}) d x$

$\begin{array}{l} \int x^{2} - \frac{x^{2}}{x^{2}} d x \\ = \int x^{2} - 1 d x = \int x^{2} d x - \int 1 d x \end{array}$

$= \frac{x^{3}}{3} - x + C$

Kindly Consider the following

8. $\int (a x^{2} + b x + c) d x$

$\begin{array}{l} = a . \int x^{2} . d x + b \int x . d x + c \int 1 . d x \\ = \frac{a \cdot x^{3}}{3} + b \frac{x^{2}}{2} + c x + d . \end{array}$

Kindly Consider the following

9. $\int (2 x^{2} + e^{x}) d x$

$\int 2 x^{2} + e^{x} d x$

$\begin{array}{l} = 2 \cdot \int x^{2} d x + \int e^{x} d x \\ = \frac{2 \cdot x^{3}}{3} + e^{x} + C \end{array}$

Kindly Consider the following

Kindly go through the solution

Kindly Consider the following

11. $\int \frac{x^{3} + 5 x^{2} - 4}{x^{2}} d x$

$\begin{array}{l} \int (\frac{x^{3}}{x^{2}} + \frac{5 x^{2}}{x^{2}} - \frac{4}{x^{2}}) d x \\ \begin{array}{l} = \int (x + 5 - 4 x^{- 2}) d x \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} = \int x . d x + \int 5 d x - \int 4 x^{- 2} d x \end{array} \\ = \frac{x^{2}}{2} + 5 x - 4 \frac{x^{- 2 + 1}}{- 2 + 1} \\ = \frac{x^{2}}{2} + 5 x - \frac{4 x^{- 1}}{- 1} \\ = \frac{x^{2}}{2} + 5 x + \frac{4}{x} + C \end{array}$

Kindly Consider the following

13. $\int \frac{x^{3} - x^{2} + x - 1}{x - 1} d x$

$\begin{array}{l} \int \frac{x^{3} - x^{2}}{x - 1} + \frac{x - 1}{x - 1} d x \\ \int (x^{2} \frac{(x - 1)}{x - 1} + 1) d x \\ \begin{array}{l} = \int (x^{2} + 1) d x \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} = \int x^{2} . d x + \int 1 . d x \end{array} \\ = \frac{x^{3}}{3} + x + C \end{array}$

Kindly Consider the following

14. $\int (1 - x)\sqrtx d x$

$\begin{array}{l} \int (\sqrtx - x \cdot x^{1 / 2}) d x \\ \int (\sqrtx - x^{3 / 2}) \cdot d x \\ \int x^{1 / 2} \cdot d x - \int x^{3 / 2} \cdot d x \\ \frac{x^{3 / 2}}{3 / 2} - \frac{x^{5 / 2}}{5 / 2} + C \\ \frac{2}{3} x^{3 / 2} - \frac{2}{5} x^{5 / 2} + \end{array}$ c

Kindly Consider the following

15. $\int\sqrtx (3 x^{2} + 2 x + 3) d x$

$\begin{array}{l} \begin{array}{l} = \int (3 x^{2} . x^{1 / 2} + 2 x . x^{1 / 2} + 3 x^{1 / 2}) . d z \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} = \int (3 x^{5 / 2} + 2 x^{3 / 2} + 3 x^{4 / 2}) . d x \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} = \int 3 x^{5 / 2} . d x + \int 2 x^{3 / 2} . d x + \int 3 x^{1 / 2} . d x \end{array} \\ = 3 \cdot \frac{x^{7 / 2}}{7 / 2} + \frac{2 \cdot x^{5 / 2}}{5 / 2} + \frac{3 \cdot x^{3 / 2}}{3 / 2} + C \\ = \frac{6 \cdot x^{7 / 2}}{7} + \frac{4 x^{5 / 2}}{5} + \frac{6}{3} x^{3 / 2} + C \\ = \frac{6 x^{7 / 2}}{7} + \frac{4 \cdot x^{5 / 2}}{5} + 2 x^{3 / 2} + C . \end{array}$

Kindly Consider the following

16. $\int (2 x - 3 c o s x + e^{x}) d x$

$= \int 2 x d x - \int 3 c o s x + e^{x} d x$

$\begin{array}{l} = \frac{2 x^{2}}{2} - 3 s i n x + e^{x} + C \\ = x^{2} - 3 s i n x + e^{x} + C \end{array}$

Kindly Consider the following

17. $\int (2 x^{2} - 3 s i n x + 5√x) d x$

$\begin{array}{l} = \int 2 x^{2} . d x - \int 3 s i n x . d x + \int 5√x d x \\ = \frac{2 x^{3}}{3} + 3 c o s x + \frac{5 x^{3 / 2}}{\frac{3}{2}} + C \\ = \frac{2 x^{3}}{3} + 3 c o s x + \frac{10 x^{3 / 2}}{3} + C \end{array}$

Kindly Consider the following

18. $\int s e c x (s e c x + t a n x) d x$

$= \int (s e c^{2} x + s e c x t a n x) d x$

$= \int s e c^{2} x . d x + \int s e c x t a n x d x$

$= t a n x + s e c x + C$

Kindly Consider the following

19. $\int \frac{s e c^{2} x}{c o s e c^{2} x} d x$

$\begin{array}{l} \int \frac{\frac{1}{c o s^{2}} x}{\frac{1}{c o s^{2}} x} d x \\ = \int \frac{s i n^{2} x}{c o s^{2} x} d x \\ \begin{array}{l} = {\int t a n}^{2} x d x \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} = \int (s e c^{2} x - 1) d x = \int s e c^{2} x d x - \int 1 . d x \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} = t a n x - x + C \end{array} \end{array}$

Kindly Consider the following

20. $\int \frac{2 - 3 s i n x}{c o s^{2} x} d x$

$\begin{array}{l} \int (\frac{2}{c o s^{2} x} - \frac{3 s i n x}{c o s^{2} x}) d x \\ \begin{array}{l} = \int (2 s e c^{2} x - 3 t a n x . s e c x) d x \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} = \int 2 s e c^{2} x . d x - \int 3 t a n x . s e c x . d x \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} = 2 t a n x - 3 s e c x + C \end{array} \end{array}$

Choose the correct answer in Exercises 21 and 22.

$\frac{x^{3 / 2}}{\frac{3}{2}} + \frac{x^{1 / 2}}{\frac{1}{2}} + C$

$= \frac{2}{3} x^{3 / 2} + 2 x^{1 / 2} +$ C

? The correct Answer is (C)

Kindly Consider the following

22. $\frac{d}{d x} f (x) = 4 x^{3} - \frac{3}{x^{4}}, f (2) = 0$

$\begin{array}{l} f (x) = \int 4 x^{3} - \frac{3}{x^{4}} d x \\ = 4 \cdot \frac{x^{4}}{4} - 3 \cdot \frac{x^{- 3}}{- 3} + C \\ = x^{4} + \frac{1}{x^{3}} + C \end{array}$

Now, $f (2) = 0 ∴ f (2) = 2^{4} + \frac{1}{2^{3}} + C = 0$

$\begin{array}{l} = 16 + \frac{1}{8} + c = 0 \\ = c = - (16 + \frac{1}{8}) \\ = c = - \frac{129}{8} \end{array}$

Therefore, correct answer is A.

Kindly Consider the following

23. $\frac{2 x}{1 + x^{2}}$

$\begin{array}{l} L e t, 1 + x^{2} = t \\ \Rightarrow 2 x d x = d t \\ I = \int \frac{2 x}{1 + x^{2}} d x = \int \frac{1 d t}{t} \\ \Rightarrow l o g | t | + c \\ \Rightarrow l o g | 1 + x^{2} | + c \end{array}$

Kindly Consider the following

25. $\frac{1}{x + x l o g x}$

$\begin{array}{l} x (1 + l o g x) \\ P u t, 1 + l o g x = t \\ = \frac{1 d x}{x} = d t \\ = I = \int \frac{1}{x + x l o g x} d x = \int \frac{1}{t} d t \\ = l o g (t) + c \\ = l o g (1 + l o g x) + c \end{array}$

Kindly Consider the following

26. $s i n x s i n (c o s x)$

Kindly go through the solution

Kindly Consider the following

$\begin{array}{l} P u t, 1 + 2 x^{2} = t \\ d x 2.2 x = d t \\ d x . 4 x = d t \end{array}$

Kindly Consider the following

31.

Kindly go through the solution

Kindly Consider the following

33.

Kindly go through the solution

Kindly Consider the following

34. ${(x^{3} - 1)}^{\frac{1}{3}} x^{5}$

$\begin{array}{l} P u t, x^{3} - 1 = t \\ 3 x^{2} d x = d t \\ = I \int {(x^{3} - 1)}^{\frac{1}{3}} x^{5} d x \\ = \int {(x^{3} - 1)}^{\frac{1}{3}} \cdot x^{3} \cdot x^{2} \cdot d x \\ = \int t^{\frac{1}{3}} (t + 1) \frac{d t}{3} = \frac{1}{3} \int (t^{\frac{4}{3}} + t^{\frac{1}{3}}) d t \\ = \frac{1}{3} [\frac{t^{\frac{7}{3}}}{\frac{7}{3}} + \frac{t^{4 / 3}}{\frac{4}{3}}] + C \\ = \frac{1}{3} [\frac{3}{7} t^{\frac{7}{3}} + \frac{3}{4} t^{\frac{4}{3}}] + C \\ = \frac{1}{7} {(x^{3} - 1)}^{\frac{7}{3}} + \frac{1}{4} {(x^{3} - 1)}^{\frac{4}{3}} + C \end{array}$

Kindly Consider the following

35. $\frac{x^{3}}{{(2 + 3 x^{3})}^{3}}$

$\begin{array}{l} P u t, 2 + 3 x^{3} = t \\ 9 x^{2} d x = d t \\ I = \int \frac{x^{3}}{{(2 + 3 x^{3})}^{3}} d x = \frac{1}{9} \int \frac{d t}{t^{3}} \\ = \frac{1}{9} [\frac{t^{- 2}}{- 2}] + C \\ = \frac{- 1}{18} (\frac{1}{t^{2}}) + C = \frac{- 1}{18 {(2 + 3 x^{3})}^{2}} + C \end{array}$

Kindly Consider the following

37. $\frac{x}{9 - 4 x^{2}}$

$\begin{array}{l} P u t 9 - 4 x^{2} = t \\ - 8 x d x = d t \\ x d x = \frac{- 1}{8} d t \\ - \frac{1}{8} \int \frac{d t}{t} = - \frac{1}{8} l o g | t | + c = - \frac{1}{8} l o g | 9 - 4 x^{2} | + c \end{array}$

Kindly Consider the following

38. $e^{2 x + 3}$

$\begin{array}{l} P u t 2 x + 3 = t \\ 2 d x = d t \\ I = \int e^{2 x + 3} d x \\ = \frac{1}{2} \int e^{t} \cdot d t \\ = \frac{1}{2} (e^{t}) + C \\ = \frac{1}{2} e^{(2 x + 3)} + C \end{array}$

Kindly Consider the following

39. $\frac{x}{e^{x^{2}}}$

$\begin{array}{l} P u t x^{2} = t \\ 2 x d x = d t \\ I = \int \frac{x}{e^{x^{2}}} d x = \frac{1}{2} \int \frac{1}{e^{t}} d t \\ = \frac{1}{2} (\frac{e^{- t}}{- 1}) + C = \frac{- 1}{2} e^{- x^{2}} + C = \frac{- 1}{2 e^{x^{2}}} + C \end{array}$

Kindly Consider the following

40. $\frac{e^{t a n^{- 1} x}}{1 + x^{2}}$

$\begin{array}{l} P u t t a n^{- 1} x = t \\ \frac{1 d x}{1 + x^{2}} = d t \\ I = \int \frac{e^{t a n^{- 1} x}}{1 + x^{2}} d x = \int e^{t} d t = e^{t} + c \\ = e^{t a n^{- 1} x} + c \end{array}$

Kindly Consider the following

43. $t a n^{2} (2 x - 3)$

$\begin{array}{l} P u t 2 x - 3 = t \\ 2 d x = d t \\ I = \int t a n^{2} (2 x - 3) d x = \int [s e c^{2} (2 x - 3) - 1] d x \\ = \frac{1}{2} \int (s e c^{2} t) d t - \int 1 d x \\ = \frac{1}{2} t a n t - x + C = \frac{1}{2} t a n (2 x - 3) - x + C \end{array}$

Kindly Consider the following

44. $s e c^{2} (7 - 4 x)$

$\begin{array}{l} P u t 7 - 4 x = t \\ - 4 d x = d t \\ I = \int s e c^{2} (7 - 4 x) d x \\ = \frac{- 1}{4} \int s e c^{2} t d t = \frac{- 1}{4} (t a n t) + C \end{array}$

Kindly Consider the following

45.

Kindly go through the solution

Kindly Consider the following

46. $\frac{2 c o s x - 3 s i n x}{6 c o s x + 4 s i n x}$

$\begin{array}{l} \frac{2 c o s x - 3 s i n x}{2 (3 c o s x + 2 s i n x)} \\ P u t 3 c o s x + 2 s i n x = t \\ \Rightarrow (- 3 s i n x + 2 c o s x) d x = d t \\ = \int \frac{2 c o s x - 3 s i n x}{6 c o s x + 4 s i n x} d x \\ = \int \frac{d t}{2 t} = \frac{1}{2} \int \frac{1}{t} d t \\ = \frac{1}{2} l o g | t | + C \\ = \frac{1}{2} l o g | 3 c o s x + 2 s i n x | + C . \end{array}$

Kindly Consider the following

47. $\frac{1}{c o s^{2} x {(1 - t a n x)}^{2}}$

$\begin{array}{l} \frac{s e c^{2} x}{{(1 - t a n x)}^{2}} \\ P u t (1 - t a n x) = t \\ \Rightarrow s e c^{2} x d x = d t \\ I = \int \frac{s e c^{2} x}{{(1 - t a n x)}^{2}} d x = \int \frac{- d t}{t^{2}} \\ = - \int t^{- 2} d t \\ = \frac{1}{t} + C \\ = \frac{1}{1 - t a n x} + C \end{array}$

Kindly Consider the following

48.

Kindly go through the solution

Kindly Consider the following

49.

Kindly go through the solution

Kindly Consider the following

50.

Kindly go through the solution

Kindly Consider the following

51. $c o t x l o g s i n x$

$\begin{array}{l} P u t l o g s i n x = t \\ \Rightarrow \frac{1}{s i n x} \cdot c o s x d x = d t \\ \Rightarrow c o t x d x = t x \\ I = \int c o t x l o g s i n x d x \\ = \int t d t = \frac{t^{2}}{2} + C = \frac{{(l o g s i n x)}^{2}}{2} + C \end{array}$

Kindly Consider the following

52. $\frac{s i n x}{1 + c o s x}$

$\begin{array}{l} P u t 1 + c o s x = t \\ \Rightarrow - s i n x d x = d t \\ I = \int \frac{s i n x}{1 + c o s x} d x = \int - \frac{d t}{t} \\ = - l o g | t | + c \\ = - l o g | 1 + c o s x | + c \end{array}$

Kindly Consider the following

53. $\frac{s i n x}{{(1 + c o s x)}^{2}}$

$\begin{array}{l} P u t 1 + c o s x = t \\ - s i n x d x = d t \\ I = \int \frac{s i n x}{{(1 + c o s x)}^{2}} d x \\ = - \int \frac{d t}{t^{2}} = - \int t^{- 2} d t \\ = \frac{1}{t} + C \\ = \frac{1}{1 + c o s x} + C \end{array}$

Kindly Consider the following

54. $\frac{1}{1 + c o t x}$

$\begin{array}{l} \int \frac{1}{1 + \frac{c o s x}{s i n x}} \cdot d x = \int \frac{1}{\frac{s i n x + c o s x}{s i n x}} d x \\ = \int \frac{s i n x}{s i n + c o s x} \cdot d x = \frac{1}{2} \int \frac{2 s i n x}{s i n x + c o s x} d x \\ = \frac{1}{2} \int (\frac{s i n x + c o s x) + (s i n x - c o s x)}{s i n x + c o s x} d x \\ = \frac{1}{2} \int 1 \cdot d x + 1 \int \frac{(s i n x - c o s x)}{s i n x + c o s x} d x \\ = \frac{1}{2} \int 1 \cdot d x + 1 \int \frac{(s i n x - c o s x)}{s i n x + c o s x} d x \\ = \frac{1}{2} (x) + \frac{1}{2} \int \frac{s i n x - c o s x}{s i n x + c o s x} d x \\ P u t s i n x + c o s x = t \\ \Rightarrow (c o s x - s i n x) d x = d t \\ = \frac{x}{2} + \frac{1}{2} \int \frac{- d t}{t} \\ = \frac{x}{2} - \frac{1}{2} l o g | t | + c \\ = \frac{x}{2} - \frac{1}{2} l o g | s i n x + c o s x | + c \end{array}$

Kindly Consider the following

55. $\frac{1}{1 - t a n x}$

$\begin{array}{l} I = \int \frac{1}{1 - t a n x} d x \\ = \int \frac{1}{1 - \frac{s i n x}{c o s x}} d x = \int \frac{1}{\frac{c o s x - s i n x}{c o s x}} d x \\ = \int \frac{c o s x}{c o s x - s i n x} d x \\ = \frac{1}{2} \int \frac{2 c o s x}{c o s x - s i n x} d x \\ = \frac{1}{2} \int \frac{(c o s x - s i n x) + (c o s x + s i n x)}{(c o s x - s i n x)} d x \\ = \frac{1}{2} \int 1 \cdot d x + \frac{1}{2} \int \frac{c o s x + s i n x}{c o s x - s i n x} d x \\ P u t c o s x - s i n x = t \\ \Rightarrow (- s i n x - c o s x) d x = d t \\ I = \frac{x}{2} + \frac{1}{2} \int \frac{- d t}{t} \\ = \frac{x}{2} - \frac{1}{2} l o g | c o s x - s i n x | + C \end{array}$

Kindly Consider the following

56.

Kindly go through the solution

Kindly Consider the following

57. $\frac{{(1 + l o g x)}^{2}}{x}$

$\begin{array}{l} P u t (1 + l o g x) = t \\ \Rightarrow \frac{1}{x} d x = d t \\ I = \int \frac{{(1 + l o g x)}^{2}}{x} d x \\ = \int t^{2} d t \\ = \frac{t^{3}}{3} + C \frac{{(1 + l o g x)}^{3}}{3} + C \end{array}$

Kindly Consider the following

58. $\frac{(x + 1) {(x + l o g x)}^{2}}{x}$

$\begin{array}{l} I = \frac{(x + 1)}{x} {(x + l o g x)}^{2} \\ = (1 + \frac{1}{x}) {(x + l o g x)}^{2} \\ P u t x + l o g x = t \\ \Rightarrow 1 + \frac{1}{x} d x = d t \\ I = \int (1 + \frac{1}{x}) {(x + l o g x)}^{2} d x \\ = \int t^{2} d t = \frac{t^{3}}{3} + C \\ = \frac{{(x + l o g x)}^{3}}{3} + C \end{array}$

Kindly Consider the following

59. $\frac{x^{3} s i n (t a n^{- 1} x^{4})}{1 + x^{8}}$

$\begin{array}{l} P u t x^{4} = t \\ 4 x^{3} d x = d t \\ I = \int \frac{x^{3} s i n (t a n^{- 1} x^{4})}{1 + x^{8}} d x \\ = \frac{1}{4} \int \frac{s i n (t a n^{- 1} t)}{1 + t^{2}}_____(1) \\ P u t t a n^{- 1} t = u \\ \Rightarrow \frac{1}{1 + t^{2}} d t = d u \end{array}$

From (1), we get

$\begin{array}{l} I = \int \frac{x^{3} s i n (t a n^{- 1} x^{4})}{1 + x^{8}} d x \\ = \frac{1}{4} \int s i n u d u \\ = \frac{1}{4} (- c o s u) + C \\ = - \frac{1}{4} c o s (t a n^{- 1} t) + C \\ = - \frac{1}{4} c o s (t a n^{- 1} x^{4}) + C \end{array}$

Kindly Consider the following

155. $s i n^{- 1} (\frac{2 x}{1 + x^{2}})$

$L e t, I = \int s i n^{- 1} (\frac{2 x}{1 + x^{2}}) d x$

Putting x = tanθ tan^-1x = θ dx = sec²θdθ we get,

$I = \int s i n^{- 1} [\frac{2 t a n θ}{1 + t a n^{2} θ}] \cdot s e c^{2} θ d θ$

Kindly Consider the following

62. $s i n^{2} (2 x + 5)$

$\begin{array}{l} H e r e, \\ s i n^{2} (2 x + 5) = \frac{1 - c o s 2 (2 x + 5)}{2} \\ = \frac{1 - c o s (4 x + 10)}{2} \\ T h e n, \\ = \int s i n^{2} (2 x + 5) d x = \int \frac{1 - c o s (4 x + 10)}{2} d x \\ = \frac{1}{2} \int 1 \cdot d x - \frac{1}{2} \int c o s (4 x + 10) d x \\ = \frac{1}{2} \cdot x - \frac{1}{2} \frac{s i n (4 x + 10)}{4} + C \\ = \frac{x}{2} - \frac{1}{8} s i n (4 x + 10) + C \end{array}$

Kindly Consider the following

63. $s i n 3 x . c o s x 4 x$

$\begin{array}{l} H e r e, \\ s i n A c o s B = \frac{1}{2} {s i n (A+B) + s i n (A -B)} \\ s i n 3 x c o s 4 x = \frac{1}{2} (s i n (3 x + 4 x) + s i n (3 x - 4 x)) \\ T h e n, \\ \int s i n 3 x c o s 4 x d x = \int \frac{1}{2} [s i n (3 x + 4 x) + s i n (3 x - 4 x) d x \\ = \frac{1}{2} \int s i n (3 x + 4 x) + s i n (3 x - 4 x) d x \\ = \frac{1}{2} \int s i n 7 x + s i n (- x) d x \\ = \frac{1}{2} [- \frac{c o s 7 x}{7} + c o s x] + c \\ = \frac{- c o s 7 x}{14} + \frac{c o s x}{2} + c \end{array}$

Kindly Consider the following

64. $c o s 2 x c o s 4 x c o s 6 x$

$\begin{array}{l} H e r e, c o s A c o s B = \frac{1}{2} {c o s (A+ B) + c o s (A- B)} \\ I = \int c o s 2 x (c o s 4 x c o s 6 x) d x \\ = \int c o s 2 x [\frac{1}{2} c o s (4 x + 6 x) + c o s (4 x - 6 x)] d x \\ = \int c o s 2 x [\frac{1}{2} (c o s 10 x + c o s (- 2 x))] d x \\ = \frac{1}{2} \int c o s 2 x c o s 10 x + c o s 2 x c o s (- 2 x) d x \\ = \frac{1}{2} \int c o s 2 x c o s 10 x + c o s 2 x d x [∴ c o s (- x) = c o s x] \\ = \frac{1}{2} [\frac{1}{2} {c o s 2 x + 10 x + c o s 2 x - 10} + {\frac{1 + c o s 4 x}{2}}] d x \\ = \frac{1}{4} \int c o s 12 x + c o s 8 x + 1 + c o s 4 x \cdot d x \\ = \frac{1}{4} [\frac{s i n 12 x}{12} + \frac{s i n 8 x}{8} + x + \frac{c o s 4 x}{4 x}] + C \end{array}$

Kindly Consider the following

65. $s i n^{3} (2 x + 1)$

$\begin{array}{l} I = \int s i n^{3} (2 x + 1) d x \\ = \int s i n^{2} (2 x + 1) . s i n (2 x + 1) d x \\ = \int {1 - c o s^{2} (2 x + 1)} s i n (2 x + 1) d x \\ P u t t i n g c o s (2 x + 1) = t \\ \Rightarrow - 2 s i n (2 x + 1) d x = d t \\ \Rightarrow s i n (2 x + 1) d x = \frac{- d t}{2} \\ I = \frac{- 1}{2} \int (1 - t^{2}) d t \\ = \frac{- 1}{2} {t - \frac{t^{3}}{3}} + C \\ = \frac{- 1}{2} {c o s (2 x + 1) - \frac{c o s^{3} (2 x + 1)^{3}}{3}} + C \\ = \frac{- c o s (2 x + 1)}{2} + \frac{c o s^{3} (2 x + 1)^{6}}{6} + C \\ = \frac{- 1}{2} c o s (2 x + 1) + \frac{1}{6} c o s^{3} (2 x + 1) + C \end{array}$

Kindly Consider the following

66. $s i n^{3} x c o s^{3} x$

$\begin{array}{l} I = \int s i n^{3} x c o s^{3} x d x \\ = \int c o s^{3} x s i n^{2} x s i n x . d x \\ = \int c o s^{3} x (1 - c o s^{2} x) . s i n x d x \\ P u t c o s x = t \\ \Rightarrow - s i n . x . d x = d t \\ I = - \int t^{3} (1 - t^{2}) d t \end{array}$

$\begin{array}{l} = - \int (t^{3} - t^{5}) d t = - {\frac{t^{4}}{4} - \frac{t^{6}}{6}} + C \\ = - {\frac{c o s^{4} x}{4} - \frac{c o s^{6} x}{6}} + C \\ = \frac{- c o s^{4} x}{4} + \frac{c o s^{6} x}{6} + C \\ = \frac{1}{6} c o s^{6} x - \frac{1}{4} c o s^{4} x + C \end{array}$

Kindly Consider the following

67. $s i n x s i n 2 x s i n 3 x$

$\begin{array}{l} H e r e, \\ s i n A s i n B = \frac{- 1}{2} {c o s (A+B) - c o s (A - B)} \\ I = \int s i n x s i n 2 x s i n 3 x \\ = \int [s i n x \cdot \frac{1}{2} {c o s (2 x - 3 x) - c o s (2 x + 3 x)}] d x \\ = \int {[s i n x \cdot \frac{1}{2} {c o s (- x) - c o s 5 x}} d x \\ = \frac{1}{2} \int s i n x c o s x - s i n x c o s 5 x d x \\ N o w, s i n 2 x = 2 s i n x c o s x, \\ = \frac{1}{2} \int \frac{s i n 2 x}{2} d x - \frac{1}{2} \int s i n x c o s 5 x d x \\ = \frac{1}{4} [\frac{- c o s 2 x}{2}] - \frac{1}{2} \int \frac{1}{2} s i n (x + 5 x) + s i n (x - 5 x) d x \\ = \frac{- c o s 2 x}{8} - \frac{1}{4} \int s i n 6 x + s i n (- 4 x) d x \\ = \frac{- c o s 2 x}{8} - \frac{1}{4} [\frac{- c o s 6 x}{6} + \frac{c o s 4 x}{4}] + C \\ = \frac{- c o s 2 x}{8} - \frac{1}{8} [\frac{- c o s 6 x}{3} + \frac{c o s 4 x}{2}] + C \\ = \frac{- c o s 2 x}{8} + \frac{c o s 6 x}{24} - \frac{c o s 4 x}{16} + C \\ = \frac{1}{4} [\frac{1}{6} c o s 6 x - \frac{c o s 4 x}{4} - \frac{c o s 2 x}{2}] + C . \end{array}$

Kindly Consider the following

68. $s i n 4 x s i n 8 x$

$\begin{array}{l} s i n A s i n B = \frac{- 1}{2} {c o s (A+B) - c o s (A-B)} \\ = \int s i n 4 x s i n 8 x d x = \int \frac{1}{2} {c o s (4 x - 8 x) - c o s (4 x + 8 x)} d x \\ = \frac{1}{2} \int c o s (- 4 x) - c o s 12 x d x \\ = \frac{1}{2} \int (c o s 4 x - c o s 12 x) d x \\ = \frac{1}{2} [\frac{s i n 4 x}{4} - \frac{s i n 12 x}{12}] + C \end{array}$

Kindly Consider the following

69. $\frac{1 - c o s x}{1 + c o s x}$

$\begin{array}{l} \frac{1 - c o s x}{1 + c o s x} = \frac{2 s i n^{2} \frac{x}{2}}{2 c o s^{2} \frac{x}{2}} \\ = t a n^{2} \frac{x}{2} = s e c^{2} \frac{x}{2} - 1 \\ = \int \frac{1 - c o s x}{1 + c o s x} d x = \int (s e c^{2} \frac{x}{2} - 1) d x \\ = [\frac{t a n \frac{x}{2}}{\frac{1}{2}}] + C \\ = 2 t a n \frac{x}{2} + C \end{array}$

Kindly Consider the following

78. $\frac{s i n^{3} x + c o s^{3} x}{s i n^{3} x \cdot c o s^{3} x}$

$\begin{array}{l} \frac{s i n^{3} x + c o s^{3} x}{s i n^{2} x \cdot c o s^{2} x} = \frac{s i n^{3} x}{s i n^{2} x \cdot c o s^{2} x} + \frac{c o s^{3} x}{s i n^{2} x . c o s^{2} x} \\ = \frac{s i n x}{c o s^{2} x} + \frac{c o s x}{s i n^{2} x} \\ = t a n x s e c x + c o t x c o s e c x . \\ = \int \frac{s i n^{3} x + c o s^{3} x}{s i n^{2} x \cdot c o s^{2} x} d x \\ \begin{array}{l} = \int (t a n x s e c x + c o t x c o s e c x) d x \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} = s e c x - c o s e c x + C \end{array} \end{array}$

Kindly Consider the following

80. $\frac{1}{s i n x c o s^{3} x}$

$\begin{array}{l} \frac{s i n^{2} x + c o s^{2} x}{s i n x c o s^{3} x} = \frac{s i n x}{c o s^{3} x} + \frac{1}{s i n x c o s x} \\ = t a n x s e c^{2} x + \frac{c o s^{2} x}{(\frac{s i n x c o s x}{c o s^{2} x})} \\ = t a n x s e c^{2} x + \frac{s e c^{2} x}{t a n x} \\ I = \int \frac{1}{s i n x c o s^{3} x} d x = \int t a n x s e c^{2} x d x + \int \frac{s e c^{2} x}{t a n x} d x \\ \begin{array}{l} P u t t a n x = t \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} S e c^{2} x d x = d t \end{array} \\ = \int \frac{1}{s i n x c o s^{3} x} d x = \int t a n x s e c^{2} x d x + \int \frac{s e c^{2} x}{t a n x} d x \\ = \int t d t + \int \frac{1 d t}{t} \\ = \frac{t^{2}}{2} + l o g | t | + C \\ = \frac{1}{2} t a n^{2} x + l o g | t a n x | + C \end{array}$

Kindly Consider the following

84. $\int \frac{s i n^{2} x - c o s^{2} x}{s i n^{2} c o s^{2} x} d x is equal to$

$\begin{array}{l} = \int \frac{s i n^{2} x - c o s^{2} x}{s i n^{2} c o s^{2} x} d x \\ \begin{array}{l} = \int (s e c^{2} x - c o s e c^{2} x) d x \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} = t a n x + c o t x + C . \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} T h e r e f o r e, t h e c o r r e c t a n s w e r i s (A) . \end{array} \end{array}$

Kindly Consider the following

86. $\frac{3 x^{2}}{x^{6} + 1}$

$\begin{array}{l} \begin{array}{l} L e t x^{3} = t \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} 3 x^{2} d x = d t \end{array} \\ I = \int \frac{3 x^{2}}{x^{6} + 1} d x = \int \frac{d t}{t^{2} + 1} \\ \begin{array}{l} = t a n^{- 1} t + C \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} = t a n^{- 1} (x^{3}) + C \end{array} \end{array}$

89. Kindly Consider the following

Kindly go through the solution

Kindly Consider the following

90. $\frac{3 x}{1 + 2 x^{4}}$

$\begin{array}{l} L e t√2 x^{2} = t \\ \Rightarrow 2√2 x d x = d t \\ I = \int \frac{3 x}{1 + 2 x^{4}} d x \\ = \frac{3}{2√2} \int \frac{d t}{1 + t^{2}} \\ = \frac{3}{2√2} [t a n^{- 1} t] + C \\ = \frac{3}{2√2} [t a n^{- 1} (\sqrt2 x^{2})] + C \end{array}$

Kindly Consider the following

91. $\frac{x^{2}}{1 - x^{6}}$

$\begin{array}{l} \begin{array}{l} L e t x^{3} = t \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} 3 x^{2} d x = d t \end{array} \\ I = \int \frac{x^{2}}{1 - x^{6}} d x = \frac{1}{3} \int \frac{d t}{1 - t^{2}} \\ = \frac{1}{3} [\frac{1}{2} l o g | \frac{1 + t}{1 - t} |] + C \\ = \frac{1}{6} l o g | \frac{1 + x^{3}}{1 - x^{3}} | + C \end{array}$

92. Kindly Consider the following

Kindly go through the solution

93. Kindly Consider the following

Kindly go through the solution

94. Kindly Consider the following

Kindly go through the solution

95. Kindly Consider the following

Kindly go through the solution

Kindly Consider the following

96. $\frac{1}{9 x^{2} + 6 x + 5}$

$\begin{array}{l} I = \int \frac{1}{9 x^{2} + 6 x + 5} d x = \int \frac{1}{{(3 x + 1)}^{2} + 2^{2}} \\ \begin{array}{l} L e t (3 x + 1) = t \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} \Rightarrow 3 d x = d t \end{array} \\ I = \int \frac{1}{{(3 x + 1)}^{2} + 2^{2}} d x \\ = \frac{1}{3} \int \frac{1}{t^{2} + 2^{2}} d t \\ = \frac{1}{3} [\frac{1}{2} t a n^{- 1} (\frac{t}{2})] + C \\ = \frac{1}{6} t a n^{- 1} (\frac{t}{2}) + C = \frac{1}{6} t a n^{- 1} (\frac{3 x + 1)}{2}) + C \end{array}$

97. Kindly Consider the following

$\begin{array}{l} 7 - 6 x - x^{2} = 7 - (x^{2} + 6 x + 9 - 9) \end{array}$

$\begin{array}{l} = 7 - (x^{2} + 6 x + 9) + 9 \end{array}$

$\begin{array}{l} = 16 - (x^{2} + 6 x + 9) \end{array}$

$\begin{array}{l} = 16 - {(x + 3)}^{2} \end{array}$

$\begin{array}{l} = (4^{2}) - {(x + 3)}^{2} \end{array}$

98. Kindly Consider the following

Kindly go through the solution

99. Kindly Consider the following

Kindly go through the solution

100. Kindly Consider the following

Kindly go through the solution

101. Kindly Consider the following

Kindly go through the solution

102. Kindly Consider the following

Kindly go through the solution

104. Kindly Consider the following

$\begin{array}{l} \frac{6 x + 7}{x^{2} - 9 x + 20} \\ L e t 6 x + 7 = A \frac{d}{d x} (x^{2} - 9 x + 20) + B \\ \Rightarrow 6 x + 7 = A (2 x - 9) + B \\ E q u a t i n g t h e c o e f f i c i e n t o f x a n d c o n s t a n t t e r m, w e h a v e \\ \begin{array}{l} 2 A = 6 a n d - 9 A + B = 7 \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} \Rightarrow A = 3 \Rightarrow B = 34 \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} ∴ 6 x + 7 = 3 (2 x - 9) + 34 \end{array} \end{array}$

105. Kindly Consider the following

Kindly go through the solution

106. Kindly Consider the following

Kindly go through the solution

Kindly Consider the following

107. $\frac{x + 3}{x^{2} - 2 x - 5}$

$\begin{array}{l} L e t (x + 3) = A \frac{d}{d x} (x^{2} - 2 x - 5) + B \\ \Rightarrow (x + 3) = A (2 x - 2) + B \\ E q u a t i n g t h e c o e f f i c i e n t o f x a n d c o n s t a n t t e r m o n b o t h s i d e, w e g e t \\ W e g e t 2 A = 1 a n d - 2 A + B = 3 \\ A = \frac{1}{2} B = 4 \\ T h e r e f o r e, (x + 3) = \frac{1}{2} (2 x - 2) + 4 - - - - - (1) \\ I = \int \frac{x + 3}{x^{2} - 2 x - 5} d x = \int \frac{\frac{1}{2} (2 x - 2) + 4}{x^{2} - 2 x - 5} d x \\ = \frac{1}{2} \int \frac{2 x - 2}{x^{2} - 2 x - 5} d x + 4 \int \frac{1}{x^{2} - 2 x + 5} d x \\ I = \int \frac{x + 3}{x^{2} - 2 x - 5} d x = \frac{1}{2} I_{1} + 4 I_{2} \\ I_{1} = \int \frac{2 x - 2}{x^{2} - 2 x - 5} d x \\ \begin{array}{l} L e t x^{2} - 2 x - 5 = t \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} \Rightarrow (2 x - 2) d x = d t \end{array} \\ I_{1} = \int \frac{d t}{t} = l o g | t | = l o g | x^{2} - 2 x - 5 | - - - - - (2) \\ I_{2} = \int \frac{1}{x^{2} - 2 x - 5} d x = \int \frac{1}{(x^{2} - 2 x + 1) - 6} d x = \int \frac{1}{{(x - 1)}^{2} - (\sqrt6)^{2}} d x \\ = \frac{1}{2√6} l o g (\frac{x - 1 -\sqrt6}{x - 1 +\sqrt6}) - - - - - (3) \\ S u b s t i t u t i n g (2) a n d (3) i n (1), w e g e t \\ I = \frac{1}{2} l o g | x^{2} - 2 x - 5 | + \frac{4}{2} l o g | \frac{x - 1 -\sqrt6}{x - 1 +\sqrt6} | + C \end{array}$ $\begin{array}{l} \end{array}$

108. Kindly Consider the following

Kindly go through the solution

Kindly Consider the following

109. $\int \frac{d x}{x^{2} + 2 x + 2} e q u a l s$

$\begin{array}{l} I = \int \frac{d x}{x^{2} + 2 x + 2} = \int \frac{d x}{(x^{2} + 2 x + 1) + 1} \\ = \int \frac{d x}{{(x + 1)}^{2} + 1} d x \\ \begin{array}{l} = [t a n^{- 1} (x + 1)] + C \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} H e n c e, t h e c o r r e c t A n s w e r i s (B) . \end{array} \end{array}$

110. Kindly Consider the following

Kindly go through the solution

111. $\frac{x}{(x + 1) (x + 2)} = \frac{A}{(x + 1)} + \frac{B}{(x + 2)}$

Kindly go through the solution

Kindly Consider the following

112. $\frac{1}{x^{2} - 9}$

Kindly go through the solution

Kindly Consider the following

113. $\frac{3 x - 1}{(x - 1) (x - 2) (x - 3)}$

Kindly go through the solution

$\begin{array}{l} S o, \frac{3 x - 1}{(x - 1) (x - 2) (x - 3)} = \frac{1}{(x - 1)} + \frac{(- 5)}{(x - 2)} + \frac{4}{(x - 3)} \\ T h e n, \int \frac{3 x - 1 d x}{(x - 1) (x - 2) (x - 3)} = \int \frac{d x}{x - 1} - 5 \int \frac{d x}{x - 2} + 4 \int \frac{d x}{x - 3} . \\ = l o g | x - 1 | - 5 l o g | x - 2 | + 4 l o g | x - 3 | + c \end{array}$

Kindly Consider the following

114. $\frac{x}{(x - 1) (x - 2) (x - 3)}$

Kindly go through the solution

Kindly Consider the following

115. $\frac{2 x}{x^{2} + 3 x + 2}$

$\frac{2 x}{x^{2} + 3 x + 2} = \frac{2 x}{(x + 1) (x + 2)} = \frac{A}{(x + 1)} + \frac{B}{(x + 2)}$

2x = A(x + 2) + B(x + 1).

= (A + B) x + (2A + B).

Comparing the coefficients we get,

A + B = 2 ---(1)

2A + B = 0 ---- (2).

So, Eqⁿ (2) - (1),

2A + B - (A + B) = 0 - 2

Þ A = - 2

And B = 2 - A = 2 - (-2) = 2 + 2 = 4

B = 4

$\begin{array}{l} S o, \frac{2 x}{(x + 1) (x + 2)} = \frac{- 2}{x + 1} + \frac{4}{x + 2} . \\ ∴ \int \frac{2 x d x}{(x + 1) (x + 2)} = - 2 \int \frac{d x}{x + 1} + 4 \int \frac{d x}{x + 2} \\ = - 2 l o g | x + 1 | + 4 l o g | x + 2 | + c \\ = 4 l o g | x + 2 | - 2 l o g | x + 1 | + c \end{array}$

Kindly Consider the following

116. $\frac{1 - x^{2}}{x (1 - 2 x)}$

$\frac{1 - x^{2}}{x (1 - 2 x)}$ which is not a proper fraction

So, division is missing

$\begin{array}{l} S o, \frac{1 - x^{2}}{x (1 - 2 x)} = \frac{1}{2} + \frac{(\frac{- x}{2} + 1)}{x (1 - 2 x)} \\ = \frac{1}{2} + \frac{1}{2} \frac{(2 - x)}{x (1 - 2 x)} \\ L e t \frac{2 - x}{x (1 - 2 x)} = \frac{A}{x} + \frac{B}{(1 - 2 x)} \end{array}$

=> 2 -x = A(1 - 2x) + B x.

So, - 2A + B = - 1

A = 2.

So, B = - 1 + 2A = - 1 + 2 ´ 2 = - 1 + 4 = 3

$\begin{array}{l} ∴ \int \frac{1 - x^{2} d x}{x (1 - 2 x)} = \int \frac{1}{2} d x + \int \frac{1}{2} \frac{(2 - x)}{x (1 - 2 x)} d x \\ = \frac{x}{2} + \frac{1}{2} {\int \frac{A}{x} d x + \int \frac{B}{1 - 2 x} d x} \\ = \frac{x}{2} + \frac{1}{2} {{\frac{2}{x} d x + \int \frac{3}{(1 - 2 x)} d x} \\ = \frac{x}{2} + l o g | x | + \frac{3}{2} l o g \frac{| 1 - 2 x |}{(- 2)} + c \\ = \frac{x}{2} + l o g | x | - \frac{3}{4} l o g | 1 - 2 x | + c \end{array}$

Kindly Consider the following

117. $\frac{x}{(x^{2} + 1) (x - 1)}$

$\frac{x}{(x^{2} + 1) (x - 1)} = \frac{A x + B}{x^{2} + 1} + \frac{C}{x - 1}$

=> x = (Ax + B)(x- 1) + C(x² + 1)

= Ax²- Ax + Bx- B + Cx² + C.

Comparing the co-efficients we get,

A + C = 0 ---- (1)

- A + B = 1 ---- (2)

- B + C = 0 ---- (3)

Adding (1) and (2) we get,

A + C - A + B = 0 + 1

=> B + C = 1 --- (4)

Adding (4) + (3) we get,

- B + C + B + C = 0 + 1

=> 2C = 1

And C = B from (3)

So, from Eqⁿ (1),

A = - C

$\begin{array}{l} A = - \frac{1}{2} . \\ ∴ \int \frac{x}{(x^{2} + 1) (x - 1)} d x = \int \frac{(- \frac{1}{2} x + \frac{1}{2})}{x^{2} + 1} d x + \int \frac{1}{2} d x . \\ = \frac{1}{2} \int \frac{1 - x}{x^{2} + 1} d x + \frac{1}{2} \int \frac{1}{x - 1} d x . \\ = \frac{1}{2} \int \frac{1}{x^{2} + 1} d x - \frac{1}{4} \int \frac{2 x d x}{x^{2} + 1} + \frac{1}{2} l o g | x - 1 | + c \\ = \frac{1}{2} t a n^{- 1} x - \frac{1}{4} l o g (x^{2} + 1) + \frac{1}{2} l o g | x - 1 | + c . \end{array}$

Kindly Consider the following

118. $\frac{x}{{(x - 1)}^{2} (x - 2)}$

$\frac{x}{{(x - 1)}^{2} (x - 2)} = \frac{A}{(x - 1)} + \frac{B}{{(x - 1)}^{2}} + \frac{C}{x + 2}$

= A(x- 1)(x + 2) + B(x + 2) + C(x- 1)²

= A(x² + 2x-x- 2) + B(x + 2) + C(x² + 1 - 2x)

= A(x² + x- 2) + B(x + 2) + C(x²- 2x + 1)

Comparing the co-efficient we get, A + C = 0 ¾ (1)

A + B - 2C = 1 ¾ (2)

- 2A + 2B + C = 0 ¾ (3)

EQⁿ (3) - 2 ´ eⁿQ. (2),

- 2A + 2B + C (2A + 2B - 4C) = 0 - 2 ´ 1.

=> - 4A + 5C = - 2 ¾ (4)

Eqⁿ (4) + 4 ´ Eqⁿ (1) we get,

- 4A + 5C + 4A + 4C = - 2 + 4 ´ 0

=> 9C = - 2

$\begin{array}{l} \Rightarrow C = - \frac{2}{9} \\ P u t t i n g C = - \frac{2}{9}, i n - - - - - (1) \\ P u t t i n g v a l u e o f A a n d C i n (3) w e g e t, \\ - 2 \times \frac{2}{9} + 2 B - \frac{2}{9} = 0 \\ \Rightarrow 2 B = \frac{2}{9} + \frac{4}{9} \\ \Rightarrow 2 B = \frac{6}{9} : \frac{2}{3} \\ \Rightarrow B = \frac{1}{3} . \end{array}$

Kindly Consider the following

119. $\frac{3 x + 5}{x^{3} - x^{2} - x + 1}$

Kindly go through the solution

Kindly Consider the following

120. $\frac{2 x - 3}{(x^{2} - 1) (2 x + 3)}$

Kindly go through the solution

Kindly Consider the following

121. $\frac{5 x}{(x + 1) (x^{2} - 4)}$

Kindly go through the solution

Kindly Consider the following

122. $\frac{x^{3} + x + 1}{x^{2} - 1}$

Kindly go through the solution

Kindly Consider the following

123. $\frac{2}{(1 - x) (1 + x^{2})}$

Kindly go through the solution

Kindly Consider the following

124. $\frac{3 x - 1}{{(x + 2)}^{2}}$

Kindly go through the solution

Kindly Consider the following

125. $\frac{1}{x^{4} - 1}$

Kindly go through the solution

$\begin{array}{l} \RightarrowA = \frac{1}{2} \\ A n d f r o m (1) w e g e t, - \frac{- 1}{2} \\ ∴ \int \frac{d x}{y^{2} - 1} = \frac{1}{2} \int \frac{d x}{y - 1} - \frac{1}{2} \int \frac{d x}{y + 1} \\ = \frac{1}{2} \int \frac{d x}{x^{2} - 1} - \frac{1}{2} \int \frac{d x}{x^{2} - 1} {y = x^{2}} \\ = \frac{1}{2} \times \frac{1}{2 \cdot 1} l o g | \frac{x - 1}{x + 1} | - \frac{1}{2} t a n^{- 1} x + C \\ = \frac{1}{4} l o g | \frac{x - 1}{x + 1} | - \frac{1}{2} t a n^{- 1} x + C \end{array}$

Kindly Consider the following

126. $\frac{1}{x (x^{n} + 1)}$

$\frac{1}{x (x^{n} + 1)} = \frac{x^{n - 1}}{x^{n - 1} \cdot x (x^{n} + 1)} = \frac{x^{n - 1}}{x^{n} (x^{n} + 1)}$

Putting xⁿ = t Þ n xⁿ^-¹dx = dt

$\begin{array}{l} ∴ \int \frac{d x}{x (x^{n} + 1)} = \int \frac{x^{n - 1} d x}{x^{n} (x^{n} + 1)} = \frac{1}{n} \int \frac{d t}{t (t + 1)} \\ = \frac{1}{n} \int \frac{(t + 1 - t)}{t (t + 1)} d t \\ = \frac{1}{n} \int {\frac{t + 1}{t (t + 1)} - \frac{t}{t (t + 1)}} d t . \\ = \frac{1}{n} {\int \frac{d t}{t} - \int \frac{d t}{t + 1}} \\ = \frac{1}{n} {l o g t - l o g (t + 1)} +C . \\ = \frac{1}{n} l o g ? \frac{t}{t + 1} + C \\ = \frac{1}{n} l o g | \frac{x^{n}}{x^{n} + 1} | +C . \end{array}$

Kindly Consider the following

127. $\frac{c o s x}{(1 - s i n x) (2 - s i n x)}$

Putting sin x = t cos xdx = dt

$\begin{array}{l} \int \frac{d t}{(1 - t) (2 - t)} = \int \frac{(2 - t) - (1 - t)}{(1 - t) (2 - t)} d t . \\ = \int \frac{d t}{1 - t} - \int \frac{d t}{2 - t} \\ = \frac{l o g | 1 - t |}{(- 1)} - \frac{l o g | 2 - t |}{(- 1)} + c \\ = l o g | \frac{2 - t}{1 - t} | + c \\ = l o g | \frac{2 - s i n x}{1 - s i n x} | + c \end{array}$

Kindly Consider the following

128. $\frac{(x^{2} + 1) (x^{2} + 2)}{(x^{2} + 3) (x^{2} + 4)}$

Kindly go through the solution

Kindly Consider the following

129. $\frac{2 x}{(x^{2} + 1) (x^{2} + 3)}$

Putting x² = t such that 2xdx = dt

$\begin{array}{l} \int \frac{2 x d x}{(x^{2} + 1) (x^{2} + 3)} = \int \frac{d t}{(t + 1) (t + 3)} = \frac{1}{2} \int \frac{2 d t}{(t + 1) (t + 3)} \\ = \frac{1}{2} \int \frac{(t + 3) - (t + 1) d t}{(t + 1) (t + 3)} \\ = \frac{1}{2} {\int \frac{d t}{t + 1} - \frac{d t}{t + 3}} \\ = \frac{1}{2} {l o g | t + 1 | - l o g | t + 3 |} + c \\ = \frac{1}{2} l o g | \frac{t + 1}{t + 3} | + c \end{array}$

Kindly Consider the following

130. $\frac{1}{x (x^{4} - 1)} = \frac{x^{3}}{x^{4} (x^{4} - 1)}$

Putting x⁴ = tÞ 4x³dx = dt

$\begin{array}{l} ∴ \int \frac{1}{x (x^{4} - 1)} d x = \frac{1}{4} \int \frac{d t}{t (t - 1)} = \frac{1}{4} \int \frac{t - (t - 1)}{t (t - 1)} d t \\ = \frac{1}{4} {\int \frac{d t}{t - 1} - \int \frac{d t}{t}} \\ = \frac{1}{4} {l o g | t - 1 | - l o g t} + C \end{array}$

$\begin{array}{l} = \frac{1}{4} l o g | \frac{t - 1}{t} | + C \\ = \frac{1}{4} l o g | \frac{x^{4} - 1}{x^{4}} | + C \end{array}$

Kindly Consider the following

131. $\frac{1}{e^{x} - 1}$

Putting e^x = t so that e^xdx = dt=>dx = $\frac{d t}{e^{x}} = \frac{d t}{t}$

$\begin{array}{l} ∴ \int \frac{d x}{e^{x} - 1} = \int \frac{d t}{(t - 1) \times t} = \int \frac{d t}{t (t - 1)} \\ = \int \frac{t - (t - 1)}{t (t - 1)} d t \\ = \int \frac{d t}{t - 1} - \int \frac{d t}{t} \\ = l o g | t - 1 | - l o g | t | + C \\ = l o g | e^{n} - 1 | - l o g e^{x} +C . \end{array}$

Kindly Consider the following

132. $\frac{x}{(x - 1) (x - 2)} e q u a l s$

Kindly go through the solution

Kindly Consider the following

133. $\frac{1}{x (x^{2} + 1)} e q u a l s$

$\begin{array}{l} \frac{1}{x (x^{2} + 1)} = \frac{x}{x^{2} (x^{2} + 1)} \\ P u t t i n g, x^{2} = t \Rightarrow 2 x d x = d t \end{array}$

$\begin{array}{l} \int \frac{d x}{x (x^{2} + 1)} = \int \frac{x d x}{x^{2} (x^{2} + 1)} = \frac{1}{2} \int \frac{d t}{t (t + 1)} \\ = \frac{1}{2} \int \frac{(t + 1) - t}{t (t + 1)} d t \\ = \frac{1}{2} {\int \frac{d t}{t} - \int \frac{d t}{t + 1}} \\ = \frac{1}{2} [l o g | t | - l o g | t + 1 |] + C \\ = \frac{1}{2} l o g | x^{2} | - \frac{1}{2} l o g | x^{2} + 1 | + C \\ = \frac{2}{2} l o g | x | - \frac{1}{2} l o g | x^{2} + 1 | + c \\ = l o g | x | - \frac{1}{2} l o g | x^{2} + 1 | + c \\ S o, o p t i o n (A) i s c o r r e c t . \end{array}$

Kindly Consider the following

134. $x s i n x$

$\begin{array}{l} \int x s i n x d x = x \int s i n d x - \int \frac{d}{d x} x \int s i n x d x d x \\ = x (- c o s x) - \int (- c o s x) d x \end{array}$

Kindly Consider the following

135. $x s i n 3 x$

$\begin{array}{l} \int x s i n 3 x d x = x \int s i n 3 x d x - \int \frac{d x}{d x} \int s i n 3 x d x d x . \\ = - x \frac{c o s 3 x}{3} + \int \frac{c o s 3 x}{3} \\ = \frac{- x c o s 3 x}{3} + \frac{s i n 3 x}{3 \times 3} + c \\ = - \frac{x}{3} c o s 3 x + \frac{1}{9} s i n 3 x + c \end{array}$

Kindly Consider the following

136. $x^{2} e^{x}$

$\begin{array}{l} \int x^{2} e^{x} d x = x^{2} \int e^{x} d x - \int \frac{d x^{2}}{d x} \int e^{x} d x d x \\ = x^{2} \cdot e^{x} - \int 2 x e^{x} d x \\ = x^{2} e^{x} - 2 [x \int e^{x} d x - \int \frac{d x}{d x} \int e^{x} d x d x] \\ = x^{2} e^{x} - 2 [x e^{x} - \int e^{x} d x] \\ = x^{2} e^{x} - 2 x e^{x} + 2 e^{x} + c \\ = e^{x} [x^{2} - 2 x + 2] + c \end{array}$

Kindly Consider the following

137. $x l o g x$

$\begin{array}{l} \int x l o g x d x = l o g x \int x d x - \int \frac{d}{d x} l o g x \cdot \int x d x d x \\ = l o g x \times \frac{x^{2}}{2} - \int \frac{1}{x} \times \frac{x^{2}}{2} d x \\ = \frac{x^{2}}{2} \cdot l o g x - \frac{1}{2} \int x d x \\ = \frac{x^{2}}{2} l o g x - \frac{1}{2} \times \frac{x^{2}}{2} + c \\ = \frac{x^{2}}{2} l o g x - \frac{x^{2}}{4} + c . \end{array}$

Kindly Consider the following

138. $x l o g 2 x$

$\begin{array}{l} \int x l o g 2 x d x = l o g 2 x \cdot \int x d x - \int \frac{d}{d x} l o g 2 x \int x d x d x \\ = \frac{x^{2}}{2} \cdot l o g 2 x - \int \frac{1}{2 x} \times \frac{d (2 x)}{d x} \times \frac{x^{2}}{2} d x + C \\ = \frac{x^{2}}{2} \cdot l o g 2 x - \int \frac{1}{2 x} \times 2 \times \frac{x^{2}}{2} 1^{d x} + C \\ = \frac{x^{2}}{2} l o g 2 x - \frac{1}{2} \int x d x + c \\ = \frac{x^{2}}{2} l o g 2 x - \frac{1}{2} \cdot \frac{x^{2}}{2} + c \\ = \frac{x^{2}}{2} l o g 2 x - \frac{x^{2}}{4} + c \end{array}$

Kindly Consider the following

139. $x^{2} \cdot l o g x$

$\begin{array}{l} \int x^{2} \cdot l o g x \cdot d x = l o g x \cdot \int x^{2} d x - \int \frac{d}{d x} l o g x \cdot \int x^{2} d x d x \\ = l o g x \cdot \frac{x^{3}}{3} - \int \frac{1}{x} \cdot \frac{x^{3}}{3} d x \\ = \frac{x^{3}}{3} \cdot l o g x - \int \frac{x^{2}}{3} d x \\ = \frac{x^{3}}{3} \cdot l o g x - \frac{1}{3} \times \frac{x^{3}}{3} + c \\ = \frac{x^{3}}{3} l o g x - \frac{x^{3}}{9} + c . \end{array}$

Kindly Consider the following

140. $\int x s i n^{- 1} x$

Kindly go through the solution

Kindly Consider the following

141. $x t a n^{- 1} x$

$\begin{array}{l} \int x t a n^{- 1} x d x = t a n^{- 1} x \int x d x - \int \frac{d}{d x} t a n^{- 1} x \cdot \int x d x . d x \\ = t a n^{- 1} x \cdot \frac{x^{2}}{2} - \int \frac{1}{1 + x^{2}} \cdot \frac{x^{2}}{2} d x . \\ = \frac{x^{2}}{2} t a n^{- 1} x - \frac{1}{2} \int \frac{x^{2}}{1 + x^{2}} d x \\ = \frac{x^{2}}{2} t a n^{- 1} x - \frac{1}{2} \int \frac{(1 + x^{2}) - 1}{1 + x^{2}} d x \\ = \frac{x^{2}}{2} \cdot t a n^{- 1} x - \frac{1}{2} [\int \frac{(1 + x^{2})}{1 + x^{2}} d x - \int \frac{d x}{1 + x^{2}}] \\ = \frac{x^{2}}{2} \cdot t a n^{- 1} x - \frac{1}{2} [\int d x - t a n^{- 1} x] \\ = \frac{x^{2}}{2} t a n^{- 1} x - \frac{1}{2} [x - t a n^{- 1} x] + C . \\ = \frac{1}{2} [x^{2} t a n^{- 1} x - x + t a n^{- 1} x] + C . \\ = \frac{1}{2} [(x^{2} + 1) t a n^{- 1} x \cdot - x] + C \end{array}$

Kindly Consider the following

142. $x c o s^{- 1} x$

$L e t, I = \int x c o s^{- 1} x d x$

Putting cos^-1 x =θ=> x = cosθ=>dx = - sinθdθ.

$S o, I = \int^{?} c o s θ \times θ \cdot (- s i n θ) d θ .$

$= - \frac{1}{2} \int^{?} θ (2 s i n θ c o s θ) d θ$ {θsin 2θ = 2 sinθ cosθ}

$\begin{array}{l} = - \frac{1}{2} \int^{?} θ s i n 2 θ d θ . \\ = - \frac{1}{2} [θ \int^{?} s i n 2 θ d θ - \int^{?} \frac{d θ}{d θ} \int^{?} s i n 2 θ d θ d θ] \end{array}$

$\begin{array}{l} = - \frac{1}{2} [\frac{θ (- c o s 2 θ)}{θ} - \int^{?} (\frac{(- c o s 2 θ)}{2}) d θ] \\ = \frac{θ}{4} c o s 2 θ - \frac{1}{4} \cdot \frac{s i n 2 θ}{2} + C \\ = \frac{θ}{4} c o s 2 θ - \frac{1}{8} (2 s i n θ c o s θ) + C . \\ = \frac{θ}{4} [2 c o s^{2} θ - 1] - \frac{1}{4} s i n θ c o s θ + C . {? c o s 2 θ = 2 c o s^{2} θ - 1} . \end{array}$ $\begin{array}{l} \end{array}$

Kindly Consider the following

143. ${(s i n^{- 1} x)}^{2}$

$L e t I = \int^{?} {(s i n^{- 1} x)}^{2} d x$

Putting sin^-1x =θ=> x = sinθ, dx = cosθdθ.

$S o, I = \int^{?} θ^{2} \cdot c o s θ d θ = θ^{2} \int^{?} c o s θ d θ - \int^{?} \frac{d}{d θ} θ^{2} \int^{?} c o s θ d θ d θ .$

$= θ^{2} s i n θ - \int^{?} 2 θ s i n θ d θ . = θ^{2} s i n θ - 2 \int^{?} θ s i n θ d θ .$

$\begin{matrix} = θ^{2} s i n θ - 2 [θ \int^{?} s i n θ d θ - \int^{?} \frac{d θ}{d θ} \int^{?} s i n θ d θ d θ] \end{matrix}$

$= θ^{2} s i n θ - 2 [θ (- c o s θ) - \int^{?} (- c o s θ) d θ]$

$= θ^{2} s i n θ + 2 θ c o s θ - 2 s i n θ + C$

Kindly Consider the following

146. $t a n^{- 1} x$

$\begin{array}{l} \int t a n^{- 1} x d x = \int (t a n^{- 1} x) 1 \cdot d x . \\ = t a n^{- 1} x \int d x - \int \frac{d}{d x} t a n^{- 1} x \int d x d x \\ = x t a n^{- 1} x - \int \frac{1}{1 + x^{2}} \cdot x d x . \\ = x t a n^{- 1} x - \frac{1}{2} \int \frac{2 x}{1 + x^{2}} d x \\ = x t a n^{- 1} x - \frac{1}{2} \cdot l o g | 1 + x^{2} | + C \end{array}$

Kindly Consider the following

147. $x {(l o g x)}^{2}$

$\begin{array}{l} \int x {(l o g x)}^{2} = \cdot {(l o g x)}^{2} \int x d x - \int \frac{d}{d x} {(l o g x)}^{2} \cdot \int x d x d x \\ = {(l o g x)}^{2} \times \frac{x^{2}}{2} - \int 2 l o g x \times \frac{1}{2} \times \frac{x^{2}}{2} d x \\ = \frac{x^{2}}{2} {(l o g x)}^{2} - \int l o g x \cdot x d x \end{array}$

$\begin{array}{l} = \frac{x^{2}}{2} {(l o g x)}^{2} - [l o g x \int x d x - \int \frac{d}{d x} l o g x \int x d x d x] \\ = \frac{x^{2}}{2} {(l o g x)}^{2} - \frac{x^{2}}{2} l o g x + \int \frac{x}{2} d x \\ = \frac{x^{2}}{2} {(l o g x)}^{2} - \frac{x^{2}}{2} l o g x + \frac{x^{2}}{4} + C \end{array}$

Kindly Consider the following

148. $(x^{2} + 1) l o g x$

$\begin{array}{l} \int (x^{2} + 1) l o g x d x = l o g x \int (x^{2} + 1) d x - \int (x^{2} + 1) d x - \int \frac{d}{d x} l o g x \int (x^{2} + 1) d x d x \\ = l o g x \cdot [\frac{x^{3}}{3} + x] - \int \frac{1}{x} \times [\frac{x^{3}}{3} + x] d x \\ = [\frac{x^{3}}{3} + x] l o g x - \int [\frac{x^{2}}{3} + 1] d x \\ = [\frac{x^{3}}{3} + x] l o g x - \frac{x^{3}}{3 \times 3} - x + C \\ = [\frac{x^{3}}{3} + x] l o g x - \frac{x^{3}}{9} - x + C \end{array}$

Kindly Consider the following

149. $e^{x} (s i n x + c o s x d x)$

∴f (x) = sin x

f (x) = cos x.

e^x [f (x) + f (x)] dx = e^xf (x) + C

$\int e^{x} (s i n x + c o s x) d x = e^{x} s i n x + c$

Kindly Consider the following

150. $\frac{x e^{x}}{{(x + 1)}^{2}}$

$L e t, I = \int \frac{x e^{x}}{{(x + 1)}^{2}} d x = \int \frac{x + 1 - 1}{{(x + 1)}^{2}} e^{x} d x$

$= \int e^{x} [\frac{(x + 1)}{{(x + 1)}^{2}} + \frac{(- 1)}{{(x + 1)}^{2}}] d x .$ is of the form

e^x [f(x) + f(x)] dx

$\begin{array}{l} W h e r e, f (x) = \frac{x + 1}{{(x + 1)}^{2}} = \frac{1}{x + 1} \\ S o, f^{'} (x) = \frac{d {(x + 1)}^{- 1}}{d x} = \frac{(- 1)}{{(x + 1)}^{2}} . \\ ∴ \int \frac{x^{\cdot} \cdot e^{x}}{(1 + x^{2})} d x = e^{x} [\frac{1}{x + 1}] + C \end{array}$

Kindly Consider the following

151. $e^{x} (\frac{1 + s i n x}{1 + c o s x})$

$L e t, = \int e^{x} (\frac{1 + s i n x}{1 + c o s x}) d x .,$

$= \int e^{x} \frac{1 + 2 s i n \frac{x}{2} c o s \frac{x}{2}}{2 c o s^{2} \frac{x}{2}} d x .$ {∴ sin 2x = 2sin x cos x. cos 2x = cos^{- 2}x- 1 1 + cos 2x = 2cos²x.

$\begin{array}{l} = \int e^{x} [\frac{1}{2 c o s^{2} \frac{x}{2}} + \frac{2 s i n \frac{x}{2} c o s \frac{x}{2}}{2 c o s^{2} \frac{x}{2}}] d x] \\ = \int e^{x} [\frac{1}{2} s e c^{2} \frac{x}{2} + \frac{c o s x \cdot s i n \frac{x}{2}}{c o s \frac{x}{2}}] d x \end{array}$

$= \int e^{x} [t a n \frac{x}{2} + \frac{1}{2} s i n^{2} \frac{x}{2}] d x$ is in the form

e^x [f(x) + f(x)].dx where

$\begin{array}{l} f (x) = t a n \frac{x}{2} \\ f^{'} (x) = \frac{d}{d r} t a n \frac{x}{2} = s e c^{2} \frac{x}{2} \frac{d}{d x} (\frac{x}{2}) \\ = \frac{1}{2} s e c^{2} \frac{x}{2} \\ = e^{x} f (x) + c = e^{x} t a n \frac{x}{2} + C \end{array}$

Kindly Consider the following

152. $e^{x} (\frac{1}{x} - \frac{1}{x^{2}}) d x$

$\begin{array}{l} L e t, I = \int e^{x} (\frac{1}{x} - \frac{1}{x^{2}}) d x \int e^{x} [f (x) + f (x)] d x \\ W h e r e, f (x) = \frac{1}{x} \\ f (x) = \frac{d x}{d x} = - 1 x^{- 2} = \frac{- 1}{x^{2}} \end{array}$

So, I = e^xf (x) + C

$\begin{array}{l} = e^{x} \frac{1}{x} + c \\ = \frac{e^{x}}{x} + c \end{array}$

Kindly Consider the following

153. $\frac{x - 3 e^{x}}{{(x - 1)}^{3}}$

$\begin{array}{l} Let, I = \int \frac{x - 3}{{(x - 1)}^{3}} e^{x} d x . \\ = \int \frac{(x - 1) - 2}{{(x - 1)}^{3}} e^{x} d x . \\ = \int e^{x} [\frac{x - 1}{{(x - 1)}^{3}} - \frac{2}{{(x - 1)}^{3}}] d x \\ = \int e^{x} [\frac{1}{{(x - 1)}^{2}} + \frac{(- 2)}{{(x - 1)}^{3}}] d x, w h i c h i s i n f o r e \\ \int e^{x} [f (x) + f^{'} (x)] d x, W h e r e \\ f (x) = \frac{1}{{(x - 1)}^{2}} \\ f^{'} (x) = \frac{d}{d x} {(x - 1)}^{- 2} = \frac{- 2}{{(x - 1)}^{3}} . \\ ∴ I = e^{x} f (x) + c = e^{x} \frac{1}{{(x - 1)}^{2}} + c = \frac{e^{x}}{{(x - 1)}^{2}} + C \end{array}$

Kindly Consider the following

154. $e^{2 x} s i n x$

$\begin{array}{l} Let, I= \int e^{2 x} s i n x d x . \\ = s i n x \int e^{2 x} - \int \frac{d}{d x} s i n x \int e^{2 x} d x d x \\ = s i n x \cdot \frac{e^{2 x}}{2} - \int c o s x \cdot \frac{e^{2 x}}{2} d x \\ = \frac{e^{2 x}}{2} s i n x - \frac{1}{2} \int c o s x \cdot \times e^{2 x} d x \\ = \frac{e^{2 x}}{2} \cdot s i n x - \frac{1}{2} {c o s x \int e^{2 x} d x - \int \frac{d}{d x} c o s x \int e^{2 x} d x d x} \\ = \frac{e^{2 x}}{2} s i n x - \frac{1}{2} {c o s x \cdot \frac{e^{2 x}}{2} + \int s i n x \frac{e^{2 x}}{2} d x} \\ = \frac{e^{2 x}}{2} s i n x - \frac{e^{2 x}}{4} c o s x - \frac{1}{4} I+ C \\ \Rightarrow I + \frac{I}{4} = \frac{e^{2 x}}{2} s i n x - \frac{e^{2 x}}{4} c o s x + C \\ \Rightarrow \frac{5 I}{4} = \frac{e^{2 x}}{2} s i n x - \frac{e^{2 x}}{4} c o s x + C \\ \Rightarrow I = \frac{4}{5} [\frac{e^{2 x}}{2} s i n x - \frac{e^{2 x}}{4} c o s x] + C \\ = \frac{e^{2 x}}{5} [2 s i n x - c o s x] + C \end{array}$

Kindly Consider the following

156. $x^{2} e^{x^{3}} d x . e q u a l s$

$\begin{array}{l} L e t I = \int x^{2} {e^{x}}^{3} d x . \\ l e t, t = x^{3} \Rightarrow d t = 3 x^{2} d x \cdot \Rightarrow \frac{d t}{3} = x^{2} d x \\ S o, = \int \cdot e^{t} \frac{d t}{3} = \frac{1}{3} \int e^{t} d t . \\ = \frac{1}{3} e^{t} + C \\ = \frac{e^{x^{3}}}{3} +C . \end{array}$

So, option (A) is correct.

Kindly Consider the following

157. $e^{x} s e c x (1 + t a n x) d x$

Kindly go through the solution

158. Kindly Consider the following

Kindly go through the solution

159. Kindly Consider the following

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160. Kindly Consider the following

Kindly go through the solution

161. Kindly Consider the following

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162. Kindly Consider the following

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163. Kindly Consider the following

Kindly go through the solution

164. Kindly Consider the following

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165. Kindly Consider the following

Kindly go through the solution

166. Kindly Consider the following

Kindly go through the solution

167. Kindly Consider the following

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168. Kindly Consider the following

Kindly go through the solution

Kindly Consider the following

169. $\int_{a}^{b} x d x$

We know that $\int_{a}^{b} f (x) d x = \underset{\begin{array}{l} h \to 0 \\ n \to \infty \end{array}}{l i m} h [f (a) + f (a + h) + f (a + 2 h) + . . . . . . + f (a + (n - 1) h)]$

where $n h = b - a$

Here, a = a, b= b and f(x) = x

$∴ \int_{a}^{b} x d x = \underset{h \to 0}{l i m} h [a (a + h) + (a + 2 h) + . . . . . . + (a + (n - 1) h)]$

$\Rightarrow \int_{a}^{b} x d x = \underset{h \to 0}{l i m} h [n a + (1 + 2 + 3 + . . . . . + (n - 1))]$

$\Rightarrow \int_{a}^{b} x d x = \underset{h \to 0}{l i m} h [a n h + h \frac{n (n - 1)}{2}]$

$= \underset{h \to 0}{l i m} h [a n h + \frac{n h (n h - h)}{2}]$

$= \underset{h \to 0}{l i m} h [a (b - a) + \frac{(b - a) (b - a - h)}{2}] [? n h = b - a]$

$\begin{array}{l} = [a (b - a) + \frac{(b - a) (b - a)}{2}] \\ = (b - a) [a + \frac{b - a}{2}] \\ = (b - a) [\frac{2 a + b - a}{2}] \\ = \frac{(b - a) (b + a)}{2} \\ = \frac{b^{2} - a^{2}}{2} \end{array}$

Kindly Consider the following

170. $\int_{0}^{5} (x + 1) d x$

We know that $\int_{a}^{b} f (x) d x = \underset{\begin{array}{l} h \to 0 \\ n \to \infty \end{array}}{l i m} h [f (a) + f (a + h) + f (a + 2 h) + . . . . . . + f (a + (n - 1) h)]$

where nh = b - a

Here, a = 0, b = 5, nh = 5 and f(x) = x + 1

$\begin{array}{l} ∴ \int_{0}^{5} (x + 1) d x = \underset{\begin{array}{l} h \to 0 \\ n \to \infty \end{array}}{l i m} h [1 + (h + 1) + (2 h + 1) + . . . . . + ((n - 1) h + 1)] \\ \Rightarrow \int_{0}^{5} (x + 1) d x = \underset{\begin{array}{l} h \to 0 \\ n \to \infty \end{array}}{l i m} h [n + h (1 + 2 + 3 . . . . . + (n - 1))] \\ \Rightarrow \int_{0}^{5} (x + 1) d x = \underset{\begin{array}{l} h \to 0 \\ n \to \infty \end{array}}{l i m} h [n h + h \frac{n (n - 1)}{2}] \\ = \underset{\begin{array}{l} h \to 0 \\ n \to \infty \end{array}}{l i m} [n h + \frac{n h (n h - h)}{2}] \end{array}$

$\begin{array}{l} = \underset{h \to 0}{l i m} [5 + \frac{5 (5 - h)}{2}] \\ = [5 + \frac{5 (5 - 0)}{2}] \\ = 5 + \frac{25}{2} = \frac{35}{2} \end{array}$

Kindly Consider the following

171. $\int_{2}^{3} x^{2} d x$

We know that $\int_{a}^{b} f (x) d x = \underset{\begin{array}{l} h \to 0 \\ n \to \infty \end{array}}{l i m} h [f (a) + f (a + h) + f (a + 2 h) + . . . . . . + f (a + (n - 1) h)]$

where nh = b - a

Here, a = 2, b = 3, nh = 1 and f(x) = x²

$\begin{array}{l} ∴ \int_{a}^{b} x^{2} d x = \underset{\begin{array}{l} h \to 0 \\ n \to \infty \end{array}}{l i m} h [4 + (4 + 4 h + h^{2}) + (4 + 8 h + 2^{2} h^{2}) + . . . . . . + (4 + 4 (n - 1) h + {(n - 1)}^{2} h^{2})] \\ = \underset{\begin{array}{l} h \to 0 \\ n \to \infty \end{array}}{l i m} h [4 n + 4 h + (1 + 2 + 3 + . . . . . . + (n - 1)) + h^{2} (1^{2} + 2^{2} + . . . . . . {(n - 1)}^{2})] \\ = \underset{\begin{array}{l} h \to 0 \\ n \to \infty \end{array}}{l i m} h [4 n h + 4 h h \frac{n (n - 1)}{2} + h h h \frac{n (n - 1) (2 n - 1)}{6}] \\ = \underset{\begin{array}{l} h \to 0 \\ n \to \infty \end{array}}{l i m} h [4 n h + 4 h h \frac{(n h - h)}{2} + \frac{n h (n h - h) (2 n h - h)}{6}] \end{array}$

$\begin{array}{l} = \underset{h \to 0}{l i m} h [4 + 2 (1 - h) + \frac{(1 - h) (2 - h)}{6}] \\ = [4 + 2 (1 - 0) + \frac{(1 - 0) (2 - 0)}{6}] \\ = 6 + \frac{1}{3} = \frac{19}{3} \end{array}$

Kindly Consider the following

172. $\int_{1}^{4} (x^{2} - x) d x$

We know that $\int_{a}^{b} f (x) d x = \underset{\begin{array}{l} h \to 0 \\ n \to \infty \end{array}}{l i m} h [f (a) + f (a + h) + f (a + 2 h) + . . . . . . + f (a + (n - 1) h)]$

where nh = b - a

Here, a = 1, b = 4, nh = 3 and f(x) = x² - xf(x) = x²- x

$\begin{array}{l} ∴ \int_{1}^{4} (x^{2} + x) d x = \underset{\begin{array}{l} h \to 0 \\ n \to \infty \end{array}}{l i m} h [0 + h + h + 2 h + 4 h^{2} + . . . . . . + (n - 1) h + {(n - 1)}^{2} h^{2}] \\ = \underset{\begin{array}{l} h \to 0 \\ n \to \infty \end{array}}{l i m} h [h (1 + 2 + 3 + . . . . . . + (n - 1)) + h^{2} (1^{2} + 2^{2} + . . . . . . {(n - 1)}^{2})] \\ = \underset{\begin{array}{l} h \to 0 \\ n \to \infty \end{array}}{l i m} h [4 n h + 4 h h \frac{n (n - 1)}{2} + h h h \frac{n (n - 1) (2 n - 1)}{6}] \\ = \underset{\begin{array}{l} h \to 0 \\ n \to \infty \end{array}}{l i m} h [h (1 + 2 + 3 + . . . . . . + (n - 1)) + h^{2} (1^{2} + 2^{2} + . . . . . . {(n - 1)}^{2})] \\ = \underset{h \to 0}{l i m} h [3 \frac{(3 - h)}{2} + \frac{3 (3 - h) (2.3 - h)}{6}] \\ = [3 \frac{(3 - 0)}{2} + \frac{3 (3 - 0) (6 - 0)}{6}] \\ = [\frac{9}{2} + 9] = \frac{27}{2} \end{array}$

Kindly Consider the following

173. $\int_{- 1}^{1} e^{x} d x$

We know that $\int_{a}^{b} f (x) d x = \underset{\begin{array}{l} h \to 0 \\ n \to \infty \end{array}}{l i m} h [f (a) + f (a + h) + f (a + 2 h) + . . . . . . + f (a + (n - 1) h)]$

where nh = b - a

Here, a = -1, b = 1, nh = 2 and f(x) = e^x

$∴ \int_{- 1}^{4} e^{x} d x = \underset{\begin{array}{l} h \to 0 \\ n \to \infty \end{array}}{l i m} h [e^{- 1} + e^{- 1} e + e^{- 1} e^{2 h} + . . . . . . + e^{- 1} e^{(n - 1) h}] = \underset{\begin{array}{l} h \to 0 \\ n \to \infty \end{array}}{l i m} h e^{- 1} \frac{[{(e^{h})}^{n} - 1]}{e^{h} - 1}$

[ $?$ The series within brackets is a G.P. and $S_{n} = a \frac{r^{n} - 1}{r - 1}$ ]

$\begin{array}{l} = \underset{\begin{array}{l} h \to 0 \\ n \to \infty \end{array}}{l i m} h e \frac{(e^{n h} - 1)}{e^{h} - 1} \\ = \underset{h \to 0}{l i m} h e^{- 1} \frac{(e^{2} - 1)}{e^{h} - 1} \\ = e^{- 1} (e^{2} - 1) \underset{h \to 0}{l i m} \frac{h}{e^{h} - 1} \\ = e^{- 1} (e^{2} - 1) \times 1 [? \underset{x \to 0}{l i m} \frac{x}{e^{x} - 1} = 1] \\ = e^{- 1 + 2} - e^{- 1} = e - e^{- 1} = e - \frac{1}{e} \end{array}$

Kindly Consider the following

174. $\int_{0}^{4} (x + e^{2 x}) d x$

We know that $\int_{a}^{b} f (x) d x = \underset{\begin{array}{l} h \to 0 \\ n \to \infty \end{array}}{l i m} h [f (a) + f (a + h) + f (a + 2 h) + . . . . . . + f (a + (n - 1) h)]$

where nh = b - a

$\begin{array}{l} ∴ \int_{0}^{4} (x + e^{2 x}) d x = \underset{\begin{array}{l} h \to 0 \\ n \to \infty \end{array}}{l i m} h [1 + (h + e^{2 h}) + (2 h + e^{4 h}) + . . . . . . + ((n - 1) h + e^{2 (n - 1) h})] \\ = \underset{\begin{array}{l} h \to 0 \\ n \to \infty \end{array}}{l i m} h [(h + 2 h + . . . . . . + (n - 1) h) + (1 + e^{2 h} + e^{4 h} + . . . . . . + e^{2 (n - 1) h})] \\ = \underset{\begin{array}{l} h \to 0 \\ n \to \infty \end{array}}{l i m} h [h (1 + 2 + . . . . . . + (n - 1)) + a (\frac{r^{n} - 1}{r - 1})] \end{array}$

$= \underset{\begin{array}{l} h \to 0 \\ h \to \infty \end{array}}{l i m} [h . h \frac{n (n - 1)}{2} + \frac{1 ({(e^{2 n})}^{n} - 1)}{e^{2 h} - 1}]$

$\begin{array}{l} = \underset{\begin{array}{l} h \to 0 \\ h \to \infty \end{array}}{l i m} [\frac{n h (n h - h)}{2} + \frac{h ({(e^{2 n h})}^{n} - 1)}{e^{2 h} - 1}] \\ = \underset{\begin{array}{l} h \to 0 \\ h \to \infty \end{array}}{l i m} [\frac{4 (4 - h)}{2} + \frac{h ((e^{24}) - 1)}{e^{2 h} - 1}] \\ = [\frac{4 (4 - 0)}{2} + (e^{8} - 1) \underset{h \to 0}{l i m} \frac{h}{e^{2 h} - 1}] \\ = 8 + (e^{8} - 1) \frac{1}{2} \underset{h \to 0}{l i m} \frac{2 h}{e^{2 h} - 1} \\ = 8 + \frac{(e^{8} - 1)}{2} = \frac{e^{8} - 15}{2} \\ [? \underset{x \to 0}{l i m} \frac{x}{e^{x} - 1} = 1] \end{array}$

Kindly Consider the following

175. $\int_{- 1}^{1} (x + 1) d x$

$\begin{array}{l} \int_{- 1}^{1} (x + 1) d x = \int_{- 1}^{1} . x d x + \int_{- 1}^{1} d x \\ = {[\frac{x^{2}}{2}]}_{- 1}^{1} + {[x]}_{- 1}^{1} \\ = [\frac{1^{2}}{2} - \frac{{(- 1)}^{2}}{2}] + [1 - (- 1)] . \\ = [\frac{1}{2} - \frac{1}{2}] + [1 + 1] = 2 . \end{array}$

Kindly Consider the following

176. $\int_{2}^{3} \frac{1}{x} d x$

$\begin{array}{l} \int_{2}^{3} \frac{1}{x} d x = \int_{- 2}^{3} \frac{1}{x} d x = {[l o g x]}_{2}^{3} \\ = l o g 3 - l o g 2 \\ = l o g \frac{3}{2} \end{array}$

Kindly Consider the following

177. $\int_{1}^{2} (4 x^{3} - 5 x^{2} + 6 x + 9) d x .$

$\begin{array}{l} \int_{1}^{2} (4 x^{3} - 5 x^{2} + 6 x + 9) d x . \\ = {[4 \times \frac{x^{4}}{4} - 5 \times \frac{x^{3}}{3} + 6 \frac{x^{2}}{2} + 9 x]}_{1}^{2} \\ = {[x^{4} - \frac{5}{3} x^{3} + 3 x^{2} + 9 x]}_{1}^{2} \\ = [2^{4} - \frac{5}{x} \times 2^{3} + 3 \times 2^{2} + 9 \times 2] - [1^{4} - \frac{5}{3} \times 1^{3} + 3 \times 1^{2} + 9 \times 1] \\ = [16 - \frac{40}{3} + 12 + 18] - [1 - \frac{5}{3} + 3 + 9] \\ = [\frac{48 - 40 + 36 + 54}{3}] - [\frac{3 - 5 + 9 + 27}{3}] \\ = \frac{98}{3} - \frac{34}{3} = \frac{64}{3} . \end{array}$

Kindly Consider the following

178. $\int_{0}^{/ 4} s i n 2 x d x$

$\begin{array}{l} {\int_{0}^{/ 4} s i n 2 x d x = [\frac{- c o s 2 x}{2}]}_{0}^{π/ 4} \\ = [\frac{- c o s 2 \times π / 4}{2} + \frac{c o s 2 \times 0}{2}] \\ = \frac{- c o s π / 2}{2} + \frac{c o s 0}{2} \\ = 0 + \frac{1}{2} = \frac{1}{2} \end{array}$

Kindly Consider the following

179. $\int_{0}^{π/ 2} c o s 2 x d x$

$\begin{array}{l} {\int_{0}^{π/ 2} c o s 2 x d x = [\frac{s i n 2 x}{2}]}_{0}^{π/ 2} = \frac{s i n 2 \timesπ/2}{2} - \frac{s i n 2 \times 0}{2} \\ = \frac{s i nπ}{2} - 0 \\ = 0 \end{array}$

Kindly Consider the following

180. $\int_{4}^{5} e^{x} d x$

$\int_{4}^{5} e^{x} d x = {[e^{x}]}_{4}^{5} = e^{5} - e^{4} = e^{4} (e - 1)$

Kindly Consider the following

181. $\int_{0}^{π/ 4} t a n x d x$

Kindly go through the solution

Kindly Consider the following

182.

Kindly go through the solution

Kindly Consider the following

183.

Kindly go through the solution

Kindly Consider the following

184. $\int_{0}^{1} \frac{d x}{1 + x^{2}}$

$\begin{array}{l} 1 \frac{d x}{1 + x^{2}} = {[t a n^{- 1} x]}_{0}^{1} \\ = t a n^{- 1} (1) - t a n^{- 1} (0) \\ = π/4 - 0 \\ = π/4 \end{array}$

Kindly Consider the following

185. $\int_{2}^{3} \frac{d x}{x^{2} - 1}$

$\begin{array}{l} \int_{2}^{3} \frac{d x}{x^{2} - 1} = {[\frac{1}{2 \times 1} l o g \frac{x - 1}{x + 1}]}_{2}^{3} \\ = \frac{1}{2} l o g | \frac{3 - 1}{3 + 1} | - \frac{1}{2} l o g | \frac{2 - 1}{2 + 1} | \end{array}$

$\begin{array}{l} = \frac{1}{2} l o g \frac{2}{4} - \frac{1}{2} l o g \frac{1}{3} \\ = \frac{1}{2} [l o g \frac{1}{2} - l o g \frac{1}{3}] \\ = \frac{1}{2} l o g \frac{1 / 2}{1 / 3} \\ = \frac{1}{2} l o g \frac{3}{2 .} \end{array}$

Kindly Consider the following

186.

Kindly go through the solution

Kindly Consider the following

187. $\int_{2}^{3} \frac{x d x}{x^{2} + 1}$

$\begin{array}{l} \int_{2}^{3} \frac{x d x}{x^{2} + 1} = \frac{1}{2} \int_{2}^{3} \frac{2 x}{x^{2} + 1} d x \\ = \frac{1}{2} {[l o g | x^{2} + 1 |]}_{2}^{3} \end{array}$

$\begin{array}{l} = \frac{1}{2} l o g | 3^{2} + 1 | - \frac{1}{2} l o g | 2^{2} + 1 | \\ = \frac{1}{2} [l o g | 9 + 1 | - l o g | 4 + 1 |] \\ = \frac{1}{2} l o g \frac{10}{5} \\ = \frac{1}{2} l o g 2 . \end{array}$

Kindly Consider the following

188. $\int_{0}^{1} \frac{2 x + 3}{5 x^{2} + 1} d x$

Kindly go through the solution

Kindly Consider the following

189. $\int_{0}^{1} x e^{x^{2}} d x$

$\begin{array}{l} \int_{0}^{1} x e^{x^{2}} d x =Isay, \\ P u t t i n g, x^{2} = t \Rightarrow 2 x d x = d t \to x d x = \frac{d t}{2} \\ = \int_{0}^{1} e^{t} \frac{d t}{2} = \frac{1}{2} \int_{0}^{1} e^{t} d t \end{array}$

When,

x = 0, t = 0

x = 1, t = 1

$\begin{array}{l} = \frac{1}{2} {[e^{t}]}_{0}^{1} \\ = \frac{1}{2} [e^{1} - e^{0}] = \frac{(e - 1)}{2} \end{array}$

Kindly Consider the following

190. $\int_{1}^{2} \frac{5 x^{2}}{x^{2} + 4 x + 3}$

Kindly go through the solution

Kindly Consider the following

191.

Kindly go through the solution

Kindly Consider the following

192.

Kindly go through the solution

Kindly Consider the following

193. $\int_{0}^{2} \frac{6 x + 3}{x^{2} + 4} d x$

$\begin{array}{l} \int_{0}^{2} \frac{6 x + 3}{x^{2} + 4} d x = \int_{0}^{2} \frac{6 x}{x^{2} + 4} d x + \int_{0}^{2} \frac{3}{x^{2} + 4} d x \\ = 3 \int_{0}^{2} \frac{2 x}{x^{2} + 4} d x + 3 \int_{0}^{2} \frac{1}{x^{2} + 2^{2}} d x \\ = 3 {[l o g | x^{2} + 4 |]}_{0}^{2} + \frac{3}{2} {[t a n^{- 1} \frac{x}{2}]}_{0}^{2} \\ = 3 [l o g | 2^{2} + 4 | - l o g | 0^{2} + 4 |] + \frac{3}{2} [t a n^{- 1} \frac{2}{2} - t a n^{- 1} \frac{0}{2}] \\ = 3 [l o g \frac{8}{4}] + \frac{3}{2} [π/4 - 0] \\ = 3 l o g 2 + 3π \end{array}$

Kindly Consider the following

194. $\int_{0}^{1} (x e^{x} + s i n \frac{πx}{4}) d x$

Kindly go through the solution

Kindly Consider the following

195.

Kindly go through the solution

Kindly Consider the following

196. $\int_{0}^{2 / 3} \frac{d x}{4 + 9 x^{2}} e q u a l s$

$\int_{0}^{2 / 3} \frac{d x}{4 + 9 x^{2}} e q u a l s = \int_{0}^{2 / 3} \frac{d x}{2^{2} + {(3 x)}^{2}}$

$= \frac{1}{2} \times \frac{{[t a n^{- 1} \frac{3 x}{2}]}_{0}^{2 / 3}}{3} = \frac{1}{6} [t a n^{- 1} \frac{3}{2} \times \frac{2}{3} - t a n^{- 1} \frac{3}{2} \times 0] = \frac{1}{6} [t a n^{- 1} (1) - t a n (0)] = \frac{1}{6} [π/4 - 0] =π/24$

? Option (c) is correct.

228. $\frac{1}{x - x^{3}}$

From (2) C = - B => C = -

$\frac{1}{2} .$

So, $\frac{1}{x - x^{3}} = \frac{1}{x} + \frac{1}{2} \frac{1}{(1 - x)} - \frac{1}{2} \frac{1}{(1 + x)}$

$∴ \int \frac{d x}{x - x^{3}} = \int \frac{d x}{x} + \frac{1}{2} \int \frac{d x}{1 - x} - \frac{1}{2} \int \frac{d x}{1 + x}$

= log |x| + $\frac{1}{2}$ log $\frac{l o g | 1 - x |}{(- 1)} - \frac{1}{2}$ log |1 + x|

= $\frac{1}{2} [2 l o g | x | - l o g | 1 - x | - l o g | 1 + x |] + C$

= $\frac{1}{2} [l o g x^{2} - l o g (1 - x) (1 + x)] + C$

= $\frac{1}{2} l o g (\frac{x^{2}}{1 - x^{2}}) + C$

229. Kindly Consider the following

Kindly go through the solution

230. Kindly Consider the following

Kindly go through the solution

231. $\int \frac{d x}{x^{2} {(x^{4} + 1)}^{\frac{3}{4}}}$

Let I = $\int \frac{d x}{x^{2} {(x^{4} + 1)}^{\frac{3}{4}}}$

= $\int \frac{d x}{x^{2} {[x^{4} (1 + \frac{1}{x^{4}})]}^{\frac{3}{4}}}$ = $\int \frac{d x}{x^{2} x^{4 \times \frac{3}{4} {[1 + \frac{1}{x^{4}}]}^{\frac{3}{4}}}}$

= $\int \frac{d x}{x^{2 + 3} {[1 + \frac{1}{x^{4}}]}^{\frac{3}{4}}}$

= $\int \frac{1}{x^{5}} {[1 + \frac{1}{x^{4}}]}^{- \frac{3}{4}} d x .$

Putting $1 + \frac{1}{x^{4}} = t \Rightarrow - 4 x^{- 5} d x = d t$

$\frac{d x}{x^{5}} = \frac{d t}{- 4} .$

I = $\int t^{- \frac{3}{4}} \frac{d t}{- 4} = - \frac{1}{4} \frac{t^{- \frac{3}{4} + 1}}{- \frac{3}{4} + 1} + C$

= $\frac{1}{4} \frac{t^{\frac{1}{4}}}{\frac{1}{4}} + C$

= $t^{\frac{1}{4}} + C$

= ${[1 + \frac{1}{x^{4}}]}^{\frac{1}{4}} + C$

232. $\frac{d x}{x^{\frac{1}{2}} + x^{\frac{1}{3}}}$

Let I = $\frac{d x}{x^{\frac{1}{2}} + x^{\frac{1}{3}}} = \int \frac{d x}{x^{\frac{1}{3}} (x^{\frac{1}{6}} + 1)}$

Putting $x^{\frac{1}{6}} = t \Rightarrow$ x = t⁶. dx = 6t⁵dt.

I = $\int \frac{6 t^{5} d t}{t^{2} (t + 1)} d t$

= $6 \int \frac{t^{3}}{(t + 1)} d t .$

= $6 \int [\frac{(t^{2} - t + 1) (t + 1) - 1}{(t + 1)}] d t$

= $6 {\int (t^{2} - t + 1) d t - \int \frac{d t}{t + 1}}$

= $6 {\frac{t^{3}}{3} - \frac{t^{2}}{2} + t - l o g | t + 1 |} + C$

= $2 x^{\frac{1}{2}} - 3 t^{\frac{1}{3}} + 6 x^{\frac{1}{6}} - 6 l o g | x^{\frac{1}{6}} + 1 | + C$

233. $\frac{5 x}{(x + 1) (x^{2} + 9)}$

From (1), B = - A = $(- \frac{1}{2}) = \frac{1}{2}$

From (2) C = 5 - B = 5 $- \frac{1}{2} = \frac{10 - 1}{2} = \frac{9}{2}$

$\int \frac{5 x}{(x + 1) (x^{2} + 9)} d x = \int {\frac{- \frac{1}{2}}{x + 1} + \frac{\frac{1}{2} x + \frac{9}{2}}{x^{2} + 9}} d x$

$= - \frac{1}{2} \int \frac{d x}{x + 1} + \frac{1}{2} \int \frac{x}{x^{2} + 9} d x + \frac{9}{2} \int \frac{d x}{x^{2} + 9}$

$= - \frac{1}{2} l o g | x + 1 | + \frac{1}{4} \int \frac{9 x}{x^{2} + 9} d x + \frac{1}{2} - \int \frac{d x}{x^{2} + 3^{2}} .$

$= - \frac{1}{2} l o g | x + 1 | + \frac{1}{4} l o g (x^{2} + 9) + \frac{9}{2} \times \frac{1}{3} t a n^{- 1} \frac{x}{3} + C$

$= - \frac{1}{2} l o g | x + 1 | + \frac{1}{4} l o g (x^{2} + 9) + \frac{3}{2} t a n^{- 1} \frac{x}{3} + C$

234. $\int \frac{s i n x}{s i n (x - a)} d x$

$= \int \frac{s i n x}{s i n (x - a)} d x$

Let x- a = t=> dx = dt.

$I = \int \frac{s i n (t + a)}{s i n t} d t = \int \frac{s i n t c o s a + c o s t s i n a}{s i n t} d t$

$= c o s a \int d t + s i n a \int c o t t d t$

$= t c o s a + s i n a \cdot l o g | s i n t | + C$

$= (x - a) c o s a + s i n a l o g | s i n (x - a) | + C$

$= x c o s a + s i n a l o g | s i n (x - a) | +_{C1}$

Where C₁ = C - a cos a

235. $\int \frac{e^{5 l o g x} - e^{4 l o g x}}{e^{3 l o g x} - e^{2 l o g x}} d x$

$\int \frac{e^{5 l o g x} - e^{4 l o g x}}{e^{3 l o g x} - e^{2 l o g x}} d x = \int \frac{e^{l o g x^{5}} - e^{l o g x^{4}}}{e^{l o g x^{3}} - e^{l o g x^{2}}} d x {\begin{matrix} ? n l o g m & = l o g m^{n} \end{matrix}$

$= \int \frac{x^{5} - x^{4}}{x^{3} - x^{2}} d x {? e^{l o g x} = x$

$= \int \frac{x^{4} (x - 1)}{x^{2} (x - 1)} d x = \int x^{2} d x = \frac{x^{3}}{3} + C$

236. Kindly Consider the following

Kindly go through the solution

237. $\int \frac{s i n^{8} x - c o s^{8} x}{1 - 2 s i n^{2} x c o s^{2} x} d x$

$\int \frac{s i n^{8} x - c o s^{8} x}{1 - 2 s i n^{2} x c o s^{2} x} d x = \int \frac{(s i n^{4} x + c o s^{4} x) (s i n^{4} x - c o s^{4} x)}{(s i n^{2} x + c o s^{2} x) - s i n^{2} x c o s^{2} x - s i n^{2} x c o s^{2} x} d x$

$= \int \frac{(s i n^{4} x + c o s^{4} x) (s i n^{2} x + c o s^{2} x) (s i n^{2} x - c o s^{2} x)}{s i n^{2} x (1 - c o s^{2} x) + c o s^{2} x (1 - s i n^{2} x)} d x {\begin{matrix} ? a^{2} - b^{2} = (a + b) (a - b) & 1 = s i n^{2} x + c o s^{2} x . \end{matrix}$

$= \int^{?} \frac{(s i n^{4} x + c o s^{4} x) (s i n^{2} x + c o s^{4} x) (s i n^{2} x - c o s^{2} x) \times 1}{s i n^{2} x \cdot s i n^{2} x + c o s^{2} x \cdot c o s^{2} x} d x$

$= \int \frac{(s i n^{4} x + c o s^{2} x) (s i n^{2} x - c o s^{2} x)}{(s i n 4 x + c o s^{4} x)} d x$

$= \int - (c o s^{2} x - s i n^{2} x) d x$

$= - \int c o s 2 x d x$

$= - \frac{s i n 2 x}{2} + C$

238. $\int \frac{1}{c o s (x + a) c o s (x + b)} d x$

$I = \int \frac{1}{c o s (x + a) c o s (x + b)} d x$

$= \frac{1}{s i n (a - b)} \int \frac{s i n (a - b)}{c o s (x + a) c o s (x + b)} \cdot d x$

$= \frac{1}{s i n (a - b)} \int \frac{s i n [(x + a) - (x + b)]}{c o s (x + a) c o s (x + b)} d x$

$= \frac{1}{s i n (a - b)} \cdot \int \frac{s i n (x + a) c o s (x + b) - c o s (x + a) s i n (x + b)}{c o s (x + a) c o s (x + b)} d x$

$= \frac{1}{s i n (a - b)} [\int t a n (x + a) d x - \int t a n (x + b) d x]$

$= \frac{1}{s i n (a - b)} [- l o g | c o s (x + a) | + l o g | c o s (x + b) |] + C$

$= \frac{1}{s i n (a - b)} l o g | \frac{c o s (x + b)}{c o s (x + a)} | + C$

239. Kindly Consider the following

Kindly go through the solution

240. $\int \frac{e^{x}}{(1 + e^{x}) (2 + e^{x})} d x$

Let I $= \int \frac{e^{x}}{(1 + e^{x}) (2 + e^{x})} d x$

Putting e^x = t=>e^xdx = dt.

$= \int \frac{d t}{(1 + t) (2 + t)} = \int \frac{(t + 2) - (t + 1)}{(t + 1) (t + 2)} d t$

$= \int \frac{d t}{t + 1} - \int \frac{d t}{t + 2}$

$= l o g | t + 1 | - l o g | t + 2 | +C$

$= l o g | \frac{t + 1}{t + 2} | +C .$

$= l o g [\frac{e^{x} + 1}{e^{x} + 2}] +C .$

241. $\int \frac{1}{(x^{2} + 1) (x^{2} + 4)} d x$

Let I = $\int \frac{1}{(x^{2} + 1) (x^{2} + 4)} d x$

Putting x²= y then,

1 - A (y + 4) + 1B (y + 1)

Comparing the co-efficient,

A + B = 0 _____ (1)

4A + B = 1 ______ (2)

Equation (2) - (1),

4A + B - A - B = 1 - 0

3A = 1 =>A = $\frac{1}{3} .$

and B = - A = - $\frac{1}{3} .$

$\frac{1}{(x^{2} + 1) (x^{2} + 4)} = \frac{1}{3} \frac{1}{(y + 1)} - \frac{1}{3} \frac{1}{(y + 4)}$

$= \frac{1}{3} \frac{1}{(x^{2} + 1)} - \frac{1}{3} \frac{1}{(x^{2} + 4)}$

$∴I = \frac{1}{3} \int \frac{d x}{x^{2} + 1} - \frac{1}{3} \int \frac{d x}{x^{2} + 2^{2}}$

$= \frac{1}{3} t a n^{- 1} x - \frac{1}{6} t a n^{- 1} \frac{x}{2} + C$

242. $\int c o s^{3} x \cdot e^{l o g s i n x} d x .$

Let I = $\int c o s^{3} x \cdot e^{l o g s i n x} d x .$

$= \int c o s^{3} x s i n x d x {? e^{l o g x} = x}$

Putting cos x = t =>-sin x dx = dt =>sin x dx = -dt.

$I = \int t^{3} (- d t) = - \int t^{3} d t = - \frac{t^{4}}{4} +C = - \frac{c o s^{4} x}{4} +C .$

243. $\int e^{3 l o g x} {(x^{4} + 1)}^{- 1} d x$

Let I $= \int e^{3 l o g x} {(x^{4} + 1)}^{- 1} d x$

$= \int \frac{e^{l o g x^{3}}}{(x^{4} + 1)} d x$

$= \int \frac{x^{3}}{x^{4} + 1} d x = \frac{1}{4} \int \frac{4 x^{3}}{x^{4} + 1} d x$

$= \frac{1}{4} l o g | x^{4} + 1 | + C$

244. $\int f^{'} (a x + b) \cdot [f] {(a x + b)}^{n} d x .$

Let I $= ? f^{?} (a x + b) \cdot [f] {(a x + b)}^{n} d x .$

$= \frac{1}{a} ? [f] {(a x + b)}^{n} * [a f^{?} (a x + b)] d x .$

Putting f (ax + b) = t.

$f^{?} (a x + b) \frac{d}{d x} (a x + b) = \frac{d t}{d x}$

af' (ax + b) dx = dt

$? = \frac{1}{a} ? t^{x} d t$

$= \frac{1}{a} [\frac{t^{n + 1}}{n + 1}]$

$= \frac{1}{a (x + 1)} [f] {(a x + b)}^{n + 1} + C$

245. Kindly Consider the following

Kindly go through the solution

246. Kindly Consider the following

Kindly go through the solution

247. Kindly Consider the following

Kindly go through the solution

248. $\int^{} \frac{2 + s i n 2 x}{1 + c o s 2 x} e^{x} d x \cdot$

Let I = $\int^{?} \frac{2 + s i n 2 x}{1 + c o s 2 x} e^{x} d x \cdot$

$= \int^{?} e^{x} [\frac{2 + 2 s i n x c o s x}{2 c o s^{2} 2}] d x {\begin{array}{l} ∴ s i n 2 θ = 2 s i n θ c o s θ & c o s 2 θ = 2 c o s^{2} θ - 1 \end{array}}$

$= \int^{?} e^{x} [\frac{1}{c o s^{2} x} + \frac{s i n x}{c o s x}] d x$

$= \int^{?} e^{x} [t a n x + s e c^{2} x] d x$ is in the form

$\int^{?} e^{x} [f (x) + f^{'} (x)] d x$

$f (x) = t a n x$

$f^{'} (x) = s e c^{2} x .$

$∴ I = e^{x} f (x) + c = e^{x} t a n x + c .$

249. $\int^{} \frac{x^{2} + x + 1}{{(x + 1)}^{2} (x + 2)} d x$

Let $I = \int^{?} \frac{x^{2} + x + 1}{{(x + 1)}^{2} (x + 2)} d x$

The integrate is of form,

$\frac{x^{2} + x + 1}{{(x + 1)}^{2} (x + 2)} = \frac{A}{x + 1} + \frac{B}{{(x + 1)}^{2}} + \frac{C}{x + 2} .$

$\Rightarrow x^{2} + x + 1 = A (x + 1) (x + 2) + B (x + 2) + C {(x + 1)}^{2} .$

$= A (x^{2} + 3 x + 2) + B (x + 2) + C (x^{2} + 1 + 2 x)$

Comparing the co-efficients,

A + C = 1 ..........(1)

3A + B + C = 1 ..........(2)

2A + 2B + C = 1 .........(3)

Equation. (2) – 2 × Equation (1),

$3 A + B + 2 C - 2 A - 2 C = 1 - 2 \times 1 .$

$\Rightarrow$ A + B = –1 .......(4)

Equation (3) - (1),

2A + 2B + C = A – C = 1 – 1

$\Rightarrow$ A + 2B = 0 ........(5)

Equation (5) - Equation (4),

A + 2B - A - B = 0 - (-1)

$\Rightarrow$ B = 1.

From (4), A = -1 - B = -1 - 1 = -2.

And from (1), C = 1 - A = 1 - (–2) = 1+2=3

$\frac{x^{2} + x + 1}{{(x + 1)}^{2} (x + 2)} = \frac{- 2}{x + 1} + \frac{1}{{(x + 1)}^{2}} + \frac{3}{x + 2}$

$∴ I = \int^{?} \frac{- 2}{x + 1} d x + \int^{?} \frac{1}{{(x + 1)}^{2}} d x + \int^{?} \frac{3}{x + 2} d x$

$= - 2 l o g | x + 1 | + \int^{?} (x + 1)^{- 2} d x + 3 l o g | x + 2 | + C .$

$= - 2 l o g | x + 1 | + \frac{{(x + 1)}^{- 2 + 1}}{- 2 + 1} + 3 l o g | x + 2 | + c$

$= - 2 l o g | x + 1 | - \frac{1}{x + 1} + 3 l o g | x + 2 | + c .$

250. Kindly Consider the following

Kindly go through the solution

252. $\int_{\frac{π}{2}}^{π} e^{x} (\frac{1 - s i n x}{1 - c o s x}) d x .$

Let I = $\int_{\frac{π}{2}}^{π} e^{x} (\frac{1 - s i n x}{1 - c o s x}) d x .$

= $\int_{\frac{π}{2}}^{π} e^{x} [\frac{1 - 2 s i n \frac{x}{2} c o s \frac{x}{2}}{2 s i n^{2} \frac{x}{2}}] d x .$

= $\int_{\frac{π}{2}}^{π} e^{x} [\frac{1}{2} c o s e c^{2} \frac{x}{2} - c o t \frac{x}{2}] d x$

= – $\int_{\frac{π}{2}}^{π} e^{x} [c o t \frac{x}{2} - \frac{1}{2} c o s e c^{2} \frac{x}{2}] d x$

= $- {[e^{x} \cdot c o t \frac{x}{2}]}_{\frac{π}{2}}^{π}$ ${? \int e^{x} [f (x) + f^{'} (x)] d x = e^{n} f (x)$

= $- [e^{π} c o t \frac{x}{2} - \frac{π}{e^{2}} c o t \frac{\frac{π}{2}}{2}]$

= $- [e^{π} \times 0 - e^{\frac{π}{2}} \cdot c o t \frac{1}{4}]$

= $0 + e^{\frac{π}{2}} \times 1 .$

= $e^{\frac{π}{2}}$

253. $\int_{0}^{\frac{π}{4}} \frac{s i n x c o s x}{c o s^{4} x + s i n^{4} x} d x .$

Let I = $\int_{0}^{\frac{π}{4}} \frac{s i n x c o s x}{c o s^{4} x + s i n^{4} x} d x .$

= $\int_{0}^{\frac{π}{4}} \frac{s i n x c o s x}{c o s^{4} x (1 + \frac{s i n^{4} x}{c o s^{4} x})} d x$

= $\int_{0}^{\frac{π}{4}} \frac{\frac{s i n x}{c o s x} \cdot \frac{c o s x}{c o s x} \frac{1}{c o s^{2} x}}{(1 + t a n^{4} x)} d x$

= $\int_{0}^{\frac{π}{4}} \frac{t a n x \cdot s e c^{2} x}{1 + t a n^{4} x} d x$

= $\frac{1}{2} \int_{0}^{\frac{π}{4}} \frac{2 t a n x \cdot s e c^{2} x}{1 + t a n^{4} x} d x .$

Putting tan²x = t =>2 tan x?sec²xdx = dt

When x = 0, t = tan²x = tan² 0 = 0

x = $\frac{π}{4}$ t = tan² $\frac{π}{4}$ = 1² = 1.

? I = $\frac{1}{2} \int_{0}^{1} \frac{d t}{1 + t^{2}} = \frac{1}{2} {[t a n^{- 1} t]}_{0}^{1}$

= $\frac{1}{2} [t a n^{- 1} (1) - t a n^{- 1} (0)]$

= $\frac{1}{2} [\frac{π}{4} - 0] = \frac{π}{8}$

254. $\int_{0}^{\frac{π}{2}} \frac{c o s^{2} x}{c o s^{2} x + 4 s i n^{2} x} d x$

$= \int_{0}^{\frac{π}{2}} \frac{c o s^{2} x}{c o s^{2} x + 4 s i n^{2} x} d x$

= $\int_{0}^{\frac{π}{2}} \frac{c o s^{2} x}{c o s^{2} x + 4 (1 - c o s^{2} x)} d x {? 1 = c o s^{2} x + s i n^{2} x .$

= $\int_{0}^{\frac{π}{2}} \frac{c o s^{2} x}{c o s^{2} x + 4 - 4 c o s^{2} x} d x$

= $\int_{0}^{\frac{π}{2}} \frac{c o s^{2} x}{4 - 3 c o s^{2} x} d x$

= $- \frac{1}{3} \int_{0}^{\frac{π}{2}} \frac{- 3 c o s^{2} x}{4 - 3 c o s^{2} x} d x$

= $- \frac{1}{3} \int_{0}^{\frac{π}{2}} \frac{(4 - 3 c o s^{2} x) - 4}{4 - 3 c o s^{2} x} d x$

= $- \frac{1}{3} [\int_{0}^{\frac{π}{2}} d x - \int_{0}^{\frac{π}{2}} \frac{4}{4 - 3 c o s^{2} x} d x]$

= $- \frac{1}{3} {{[x]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} \frac{4}{4 - 3 \times \frac{1}{s e c^{2} x}} d x}$

= $- \frac{1}{3} {\frac{π}{2} - \int_{0}^{\frac{π}{2}} \frac{4 s e c^{2} x}{4 s e c^{2} x - 3} d x}$

= $- \frac{1}{3} {\frac{π}{2} - \int_{0}^{\frac{π}{2}} \frac{4 s e c^{2} x}{4 (1 + t a n^{2} x) - 3} d x} {s e n^{2} x = 1 + t a n^{2} x}$

= $\frac{- π}{6} + \frac{2}{3}_{1}$

where I₁ = $\int_{0}^{\frac{π}{2}} \frac{2 t a n^{2} x}{1 + 4 t a n^{2} x} d x .$

Putting 2tan x = t =>2 sin²xdx = dt& when x = 0, t = 2 tan (0) = 0

x = $\frac{π}{2}$ , t = 2 tan $\frac{π}{2}$ = ∞

I1 = $\int_{0}^{\infty} \frac{d t}{1 + t^{2}} .$

= $[t a n^{-} t]_{0}^{\infty}$

= tan^–1 (∞) – tan^–10

= $\frac{π}{2} - 0$ = $\frac{π}{2} .$

? I = $\frac{- π}{6} + \frac{2}{3} \times \frac{π}{2} = - \frac{π}{6} + \frac{π}{3} = \frac{π}{6} .$

255. Kindly Consider the following

Kindly go through the solution

256. Kindly Consider the following

Kindly go through the solution

257. $\int_{0}^{\frac{π}{4}} \frac{s i n x + c o s x}{9 + 16 s i n^{2} x} d x$

Let I = $\int_{0}^{\frac{π}{4}} \frac{s i n x + c o s x}{9 + 16 s i n^{2} x} d x$

Let sin x – cos x = t. =>(cosx + sin x) dx = dt.

and (sin x – cos x)²= t²

sin²x + cos²x – 2 sin x cos x = t²

1 – sin2x = t².

sin2t = 1 - t².

When x = 0, t = sin 0 – cos 0 = –1

? I = $\int_{- 1}^{0} \frac{d t}{9 + 16 (1 - t^{2})} = \int_{- 1}^{0} \frac{d t}{9 + 16 - 16 t^{2}}$

= $\int_{- 1}^{0} \frac{d t}{25 - 16 t^{2}}$

= $\int_{- 1}^{0} \frac{d t}{16 (\frac{25}{16} - t^{2})}$

= $\frac{1}{16} \int_{- 1}^{0} \frac{d t}{{(\frac{5}{4})}^{2} - t^{2}}$

= $\frac{1}{16} {[\frac{1}{2 \times (\frac{5}{4})} l o g | \frac{\frac{5}{4} + t}{\frac{5}{4} - t} |]}_{- 1}^{0} {\int \frac{d x}{a^{2} - x^{2}} = \frac{1}{2 a} l o g | \frac{a + x}{a - x} |}$

= $\frac{1}{16} \times \frac{4}{2 \times 5} {[l o g | \frac{5 + 4 t}{5 - 4 t} |]}_{- 1}^{0}$

= $\frac{1}{40} [l o g \frac{5 + 4 \times 0}{5 - 4 \times 0} - l o g \frac{5 + 4 (- 1)}{5 - 4 (- 1)}]$

= $\frac{1}{40} [l o g \frac{5}{5} - l o g \frac{1}{9}]$

= $\frac{- 1}{40} [l o g 1 - l o g 9^{(- 1)}]$

= $\frac{1}{40} [0 - (- 1) l o g 9]$

= $\frac{1}{40} l o g 9$

= $\frac{1}{40} l o g 3^{2} = \frac{2}{40} l o g 3$

= $\frac{1}{20} l o g 3$

258. $\int_{0}^{π / 2} s i n 2 x t a n^{- 1} (s i n x) d x$

Let I = $\int_{0}^{π / 2} s i n 2 x t a n^{- 1} (s i n x) d x$

= $\int_{0}^{\frac{π}{2}} 2 s i n x c o s x t a n^{- 1} (s i n x) d x$

Putting sin x = t =>cos xdx = dt.

whenx = 0, t = sin 0 = 0.

$x = π / 2 t = s i n π / 2 = 1$

? I = $\int_{0}^{1} 2 \cdot t t a n^{- 1} (t) d t$

= $2 [t a n^{-} t \int_{0}^{1} t d t - \int_{0}^{1} \frac{d}{d t} t a n^{- t} \int t d t d t]$

= $2 {{[t a n^{- 1} t \times \frac{t^{2}}{2}]}_{0}^{1} - \int_{0}^{1} \frac{1}{1 + t^{2}} \times \frac{t^{2}}{2} d t}$

= $2 {[t a n^{- 1} (1) \times \frac{1}{2} - t a n^{- 1} (0) \times \frac{0}{2}] - \frac{1}{2} \int_{0}^{1} \frac{(1 + t^{2}) - 1}{1 + t^{2}} d t}$

= $2 [\frac{π}{8} - 0] - \frac{2}{2} {\int_{0}^{1} \frac{1 + t^{2}}{1 + t^{2}} d t - \int_{0}^{1} \frac{d t}{1 + t^{2}}}$

= $\frac{π}{4} - \int_{0}^{1} d t + {[t a n^{- 1} t]}_{0}^{1}$

= $- \frac{π}{4} - {[t]}_{0}^{1} + [t a n^{- 1} (1) - t a n^{- 1} (0)]$

= $\frac{π}{4} - 1 + \frac{π}{4} = 2 \times \frac{π}{4} - 1 = \frac{π}{2} - 1$

259 $\int_{0}^{n} \frac{x t a n x}{s e c x + t a n x} d x$

$\begin{array}{l} 2 I = \int_{0}^{n} \frac{π t a n x}{s e c x + t a n x} d x \Rightarrow 2 I = π \int_{0}^{n} \frac{\frac{s i n x}{c o s x}}{\frac{1}{c o s x} + \frac{s i n x}{c o s x}} d x \\ \Rightarrow 2 I = π \int_{0}^{π} \frac{s i n x + 1 - 1}{1 + s i n x} d x \\ \Rightarrow 2 I = π \int_{0}^{π} 1 . d x - π \int_{0}^{π} \frac{1}{1 + s i n x} d x \\ \Rightarrow 2 I = π \int_{0}^{π} 1 . d x - π \int_{0}^{π} \frac{(1 - s i n x)}{(1 + s i n x) (1 - s i n x)} d x \end{array}$

$\begin{array}{l} \Rightarrow 2 I = π {[x]}_{0}^{π} - π \int_{0}^{π} \frac{1 - s i n x}{c o s^{2} x} d x \\ \Rightarrow 2 I = π^{2} - π \int_{0}^{π} (s e c^{2} x - t a n x s e c x) d x \\ \Rightarrow 2 I = π^{2} - π {[t a n x - s e c x]}_{0}^{π} \end{array}$

$\begin{array}{l} \Rightarrow 2 I = π^{2} - π [t a n π - s e c π - t a n 0 + s e c 0] \\ \Rightarrow 2 I = π^{2} - π [0 - (- 1) - 0 + 1] \\ \Rightarrow 2 I = π^{2} - 2 π \\ \Rightarrow 2 I = π (π - 2) \\ I = \frac{π}{2} (π - 2) \end{array}$

260. $\int_{1}^{4} [| x - 1 | + | x + 2 | + | x - 3 |] d x$

Let I = $\int_{1}^{4} [| x - 1 | + | x + 2 | + | x - 3 |] d x$

I = $\int_{1}^{4} | x - 1 | d x + \int_{1}^{4} | x + 2 | d x + \int_{1}^{4} | x - 3 | d x$

I = I₁ + I₂ + I₃______(1)

So, I₁ = $\int^{4}$ |x – 1| dx . ${\begin{cases} (x - 1) x - 1 > 0 \Rightarrow x > 1 \\ (x - 1) x - 1 < 0 \Rightarrow x < 1 \end{cases}$

= $\int_{1}^{4} (x - 1) d x$

= ${[\frac{x^{2}}{2} - x]}_{1}^{4} = (\frac{4^{2}}{2} - \frac{1^{2}}{2}) - (4 - 1)$

= $\frac{15}{2} - 3 = \frac{15 - 6}{2} = \frac{9}{2} .$

I₂ = $\int_{1}^{4} | x - 2 | d x$ ${\begin{cases} x - 2 x - 2 > 0, x > 2 \\ - (x - 2) x - 2 < 0, x < 2 . \end{cases}$

= $\int_{1}^{2} - (x - 2) d x + \int_{2}^{4} (x - 2) d x$

= $- {[\frac{x^{2}}{2} - 2 x]}_{1}^{2} + {[\frac{x^{2}}{2} - 2 x]}_{2}^{4}$

= $- [(\frac{2^{2}}{2} - \frac{1}{2}) - (2 \times 2 - 2 \times 1)] + [(\frac{4^{2}}{2} - \frac{2^{2}}{2}) - (2 \times 4 - 2 \times 2)]$

= $- [\frac{4 - 1}{2} - (4 - 2)] + [(8 - 2) - (8 - 4)]$

= $- \frac{3}{2} + 2 + 6 - 4$

= $\frac{- 3 + 4 + 12 - 8}{2} = \frac{5}{2}$

I₃ = $\int_{1}^{4} | x - 3 | d x$ ${\begin{cases} (x - 3) x - 3 > 0, x > 3 \\ - (x - 3) x - 3 < 0, x < 3 \end{cases}$

= $\int_{1}^{3} - (x - 3) d x + \int_{3}^{4} (x - 3) d x$

= $- {[\frac{x^{2}}{2} - 3 x]}_{1}^{3} + {[\frac{x^{2}}{2} - 3 x]}_{3}^{4}$

= $- [(\frac{3^{2}}{2} - \frac{1^{2}}{2}) - (3 \cdot 3 - 3 \cdot 1)] + [(\frac{4^{2}}{2} - \frac{3^{2}}{2}) - (3 \times 4 - 3 \times 3)]$

= $- [\frac{8}{2} - 6] + [\frac{7}{2} - 3]$

= $- 4 + 6 + \frac{7}{2} - 3 = \frac{- 8 + 12 + 7 - 6}{2} = \frac{5}{2}$

Hence Equation (1) becomes

I = $\frac{9}{2} + \frac{5}{2} + \frac{5}{2}$

I = $\frac{19}{2}$

261. $\int_{1}^{3} \frac{d x}{x^{2} (x + 1)} .$ $= \frac{2}{3} + l o g \frac{2}{3}$

Let $I = \int_{1}^{3} \frac{d x}{x^{2} (x + 1)} .$

The integrand is of the form.

1 = Ax (x + 1) + B (x + 1) + Cx²

= A (x² + x) + B (x + 1) + Cx²

Comparing the coefficients,

A + C = 0 ____ (1)

A + B = 0 ______ (2)

B = 1 ________ (3)

Putting Equation (3) in (2),

A + 1 = 0

A = -1.

and putting value of A in Equation (1),

-1 + C = 0

C = 1

$\frac{1}{x^{2} (x + 1)} = \frac{- 1}{x} + \frac{1}{x^{2}} + \frac{1}{x + 1}$

$∴ I = \int_{1}^{3} \frac{d x}{x^{2} (x + 1)} = \int_{1}^{3} \frac{- d x}{x} + \int_{1}^{3} \frac{d x}{x^{2}} + \int_{1}^{3} \frac{d x}{x + 1}$

$= - {[l o g | x |]}_{1}^{3} + {[\frac{x^{- 2 + 1}}{- 2 + 1}]}_{1}^{3} + {[l o g ? x + 1]}_{1}^{3}$

$= [- l o g 3 + l o g 1] - {[\frac{1}{x}]}_{1}^{3} + [l o g | 3 + 1 | - l o g | 1 + 1 |]$

$= - l o g 3 + 0 - [\frac{1}{3} - 1] + l o g 4 - l o g 2$

$= - l o g 3 - \frac{(1 - 3)}{3} + l o g 2^{2} - l o g 2$

$= - l o g 3 - \frac{(- 2)}{3} + 2 l o g 2 - l o g 2$

$= l o g 2 - l o g 3 + \frac{2}{3}$

$= \frac{2}{3} + l o g \frac{2}{3}$

Hence proved.

262. $\int_{0}^{1} x e^{x} d x = 1$

LHS= $\int_{0}^{1} x e^{x} d x = x \int_{0}^{1} e^{x} d x - \int_{0}^{1} \frac{d x}{d x} \int e^{x} d x d x$

$= {[x e^{x}]}_{0}^{1} - \int_{0}^{1} e^{x} d x$

$= [1 e^{1} - 0 \times e^{0}] - {[e^{x}]}_{0}^{1}$

$= (e^{1} - 0) - (e^{1} - e^{0})$

= e-e + e⁰

= e⁰ = 1=RHS

263. $\int_{- 1}^{1} x^{17} \cdot c o s^{4} x d x$ $= 0$

$= \int_{- 1}^{1} x^{17} \cdot c o s^{4} x d x$

Here f (x) = x¹⁷ cos4x

f ( -x) = ( -x)¹⁷ cos⁴ ( -x)

= -x¹⁷ cos⁴x

= f (x)

i e, odd fxⁿ

As $\int_{a}^{a} f (x) d x = 0$ for odd fxⁿ

therefore, I = 0.

264. $\int_{0}^{\frac{π}{2}} s i n^{3} x d x . = \frac{2}{3}$

LHS = I $= \int_{0}^{\frac{π}{2}} s i n^{3} x d x . {s i n 3A = 3 s i nA - 4 s i n^{3 A}$

$= \int_{0}^{\frac{π}{2}} \frac{1}{4} (3 s i n x - s i n 3 x) d x . s i n^{3A} = \frac{1}{4} (3 s i nA - s i n 3A)$

$= \frac{1}{4} [3 \int_{0}^{\frac{π}{2}} s i n x d x - \int_{0}^{\frac{π}{q}} s i n 3 x d x]$

$= \frac{1}{4} {3 {[- c o s x]}_{0}^{\frac{π}{2}} - {[\frac{- c o s 3 x}{3}]}_{0}^{\frac{π}{2}}}$

$= \frac{3}{4} (- c o s \frac{π}{2} + c o s 0) - \frac{1}{12} (- c o s \frac{3 π}{2} + c o s 3 \times 0)$

$= \frac{3}{4} (- 0 + 1) - \frac{1}{12} (- 0 + 1)$

$= \frac{3}{4} - \frac{1}{12} = \frac{9 - 1}{12} = \frac{8}{12} = \frac{2}{3}$

265. $\int_{0}^{\frac{π}{4}} \cdot 2 t a n^{3} x d x = 1 - l o g 2$

Let I $\int_{0}^{\frac{π}{4}} \cdot 2 t a n^{3} x d x$

$= 2 \int_{0}^{\frac{π}{4}} t a n^{2} x t a n x d x$

$= 2 \int_{0}^{\frac{π}{4}} (s e c^{2} x - 1) t a n x d x {? s e c^{2} x = t a n^{2} x + 1}$

$= 2 \int_{0}^{\frac{π}{4}} s e c^{2} x t a n x d x - 2 \int_{0}^{\frac{π}{4}} t a n x d x$

$= 2I_{1} + 2 [l o g] {(c o s x)}_{0}^{\frac{π}{4}}$

$= 2I_{1} + 2 [l o g (c o s \frac{π}{4}) - l o g (c o s 0)]$

$= 2I_{1} + 2 (l o g1/√2$
$\frac{}{} - l o g 1)$

$= 2I_{1} + 2 (l o g \cdot 2^{\frac{- 1}{2}} - 0)$

$I = 2I_{1} + 2 \times (\frac{- 1}{2}) l o g 2 = 2I_{1} - l o g 2 .____(1) .$

Where I $_{1} = \int_{0}^{\frac{π}{4}} s e c^{2} x t a n x d x$

Let tan x = t =>sec²xdx = dt

When, x = 0, t = tan 0. = 0

$x = \frac{π}{4}, t = t a n \frac{π}{4} = 1$

$_{1} = \int_{0}^{1} t d t = {[\frac{t^{2}}{2}]}_{0}^{1} = \frac{1}{2} - 0 = \frac{1}{2}$

So, Equation (1) becomes,

$= 2 \times \frac{1}{2} - l o g 2$

=1 - log 2.

266. $\int_{0}^{1} s i n^{- 1} x d x . = \frac{π}{2} - 1$

Kindly go through the solution

267. Evaluate $\int_{0}^{1} e^{2 - 3 x} d x$ as a limit of a sum.

Let $I = \int_{0}^{1} e^{2 - 3 x} d x$

We know that,

$\int_{a}^{b} f (x) d x = (b - a) \underset{n \to \infty}{l i m} \frac{1}{n} [f (a) + f (a + h) + . . . + f (a (n - 1) h)]$

Where, $h = \frac{b - a}{n}$

Here, $a = 0, b = 1$ and $f (x) e^{2 - 3 x}$

$\Rightarrow h = \frac{1 - 0}{n} = \frac{1}{n}$

$∴ \int_{0}^{1} e^{2 - 3 x} d x = (1 - 0) \underset{n \to \infty}{l i m} \frac{1}{n} [f (0) + f (0 + h) + . . . + f (0 + (n - 1) h)]$

$\begin{array}{l} = \underset{n \to \infty}{l i m} \frac{1}{n} [e^{2} + e^{2 - 3 x} + . . . + e^{2 - 3 (n - 1) h}] = \underset{n \to \infty}{l i m} \frac{1}{n} [e^{2} {1 + e^{- 3 h} + e^{- 6 h} + e^{- 9 h} + . . . + e^{- 3 (n - 1) h}}] \\ = \underset{n \to \infty}{l i m} \frac{1}{n} [e^{2} {\frac{1 - {(e^{- 3 h})}^{n}}{1 - (e^{- 3 h})}}] = \underset{n \to \infty}{l i m} \frac{1}{n} [e^{2} {\frac{1 - e^{\frac{- 3}{n} n}}{1 - e^{\frac{- 3}{n}}}}] \\ = \underset{n \to \infty}{l i m} \frac{1}{n} [\frac{e^{2} (1 - e^{- 3})}{1 - e^{\frac{- 3}{n}}}] = e^{2} (e^{- 3} - 1) \underset{n \to \infty}{l i m} \frac{1}{n} [\frac{1}{e^{\frac{- 3}{n}} - 1}] \\ = \underset{n \to \infty}{l i m} \frac{1}{n} [\frac{e^{2} (1 - e^{- 3})}{1 - e^{\frac{- 3}{n}}}] = e^{2} (e^{- 3} - 1) \underset{n \to \infty}{l i m} \frac{1}{n} [\frac{1}{e^{\frac{- 3}{n}} - 1}] \end{array}$

$\begin{array}{l} = e^{2} (e^{- 3} - 1) \underset{n \to \infty}{l i m} (- \frac{1}{3}) [\frac{- \frac{3}{n}}{e^{\frac{- 3}{n}} - 1}] = \frac{e^{2} (e^{- 3} - 1)}{3} \underset{n \to \infty}{l i m} [\frac{- \frac{3}{n}}{e^{\frac{- 3}{n}} - 1}] \\ = \frac{- e^{2} (e^{- 3} - 1)}{3} (1) \\ = \frac{- e^{- 1} + e^{2}}{3} \\ = \frac{1}{3} (e^{2} - \frac{1}{e}) \end{array}$

268. $\int \frac{d x}{e^{x} + e^{- x}}$ is Equal to

Let $= \int \frac{d x}{e^{x} + e^{- x}}$

$= \int \frac{d x}{e^{x} + \frac{1}{e^{x} .}} = \int \frac{e^{x} d x}{e^{x} \cdot e^{x} + 1}$

$= \int \frac{e^{x} d x}{e^{2 x} + 1}$

Putting e^x = t

e^xdx = dt.

$∴ = \int \frac{d t}{t^{2} + 1 .}$

= tan^{- 1} t + c

= tan^{- 1} (e^x) + c

therefore, Option A is correct.

269. $\int \frac{c o s 2 x d x}{{(s i n x + c o s x)}^{2}}$ is Equal to

Let $= \int \frac{c o s 2 x d x}{{(s i n x + c o s x)}^{2}}$

$= \int \frac{c o s^{2} x - s i n^{2} x}{{(s i n x + c o s x)}^{2}} d x {? c o s 2 x = c o s^{2} x - s i n^{2} x$

$= \int \frac{(c o s x + s i n x) (c o s x - s i n x)}{{(s i n x + c o s x)}^{2}} d x {? a^{2} - b^{2} = (x + b) (x - b)$

$= \int \frac{(c o s x - s i n x)}{s i n x + c o s x .} d x$

$= l o g | s i n x + c o s x | + c {\int \frac{f^{'} (x)}{f (x)} d x = l o g | f (x) | + c$

So, option B is correct.

270. If $f (a + b - x) = f (x), t h e n, \int_{a}^{b} x f (x)$ is Equal to

Giver, f(a + bx) = f(x). _________ (1)

Let I $= \int_{a}^{b} x + (x) = \int_{a}^{b} (a + b - x) f (a + b - x) d x_____(2)$

${? \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x .$

$= \int_{a}^{b} (a + b - x) f (x) d x .$ ___________ (3) {because Equation (1)}

$+ = \int_{a}^{b} [(a + b - x) f (x) + x f (x)] d x$

$\Rightarrow 2I = \int_{a}^{b} (a + b - x + x) \overset{}{f} (x) d x$

$\Rightarrow 2I = \int_{a}^{b} (a + b) f (x) d x$

$\Rightarrow = \frac{a + b}{2} \int_{a}^{b} f (x) d x .$

therefore, Option D is correct.

271. The value of $\int_{0}^{1} t a n^{- 1} (\frac{2 x - 1}{1 + x - x^{2}}) d x$ is

A.1

B. 0

C.-1

D. $\frac{π}{4}$

Consider, $I = \int_{0}^{1} t a n^{- 1} (\frac{2 x - 1}{1 + x - x^{2}}) d x$

$\begin{array}{l} \Rightarrow I = \int_{0}^{1} t a n^{- 1} (\frac{x - (1 - x)}{1 + x (1 - x)}) d x \\ \Rightarrow I = \int_{0}^{1} [t a n^{- 1} x - t a n^{- 1} (1 - x)] d x - - - - - (1) \\ \Rightarrow I = \int_{0}^{1} [t a n^{- 1} (1 - x) - t a n^{- 1} (1 - 1 + x)] d x \\ \Rightarrow I = \int_{0}^{1} [t a n^{- 1} (1 - x) - t a n^{- 1} x] d x \\ \Rightarrow I = \int_{0}^{1} [t a n^{- 1} (1 - x) - t a n^{- 1} (x)] d x - - - - - (2) \end{array}$

Adding (1) and (2), we get

$\begin{array}{l} \Rightarrow 2 I = \int_{0}^{1} [t a n^{- 1} x - t a n^{- 1} (1 - x) - t a n^{- 1} (1 - x) - t a n^{- 1} x] d x \\ \Rightarrow 2 I = 0 \\ \Rightarrow I = 0 \end{array}$

Thus, the correct option is B.

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