Maths Integrals

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

The integrate is of form,

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

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

$=A\left(x^2+3x+2\right)+B(x+2)+C\left(x^2+1+2x\right)$

Comparing the co-efficients,

A + C = 1 .(1)

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

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

Equation. (2) – 2 * Equation (1),

$3A+B+2C-2A-2C=1-2*1.$

$\Rightarrow$ 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

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

$\therefore I={{\displaystyle \int }}^{\text{?}}\frac{-2}{x+1}dx+{{\displaystyle \int }}^{\text{?}}\frac{1}{{(x+1)}^2}dx+{{\displaystyle \int }}^{\text{?}}\frac{3}{x+2}dx$

$=-2log|x+1|+{{\displaystyle \int }}^{\text{?}}(x+1){}^{-2}dx+3log|x+2|+C.$

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

$=-2log|x+1|-\frac{1}{x+1}+3log|x+2|+c.$

Let I = ${{\displaystyle \int }}^{\text{?}}\frac{2+sin2x}{1+cos2x}{e}^{x}dx\cdot$

$={{\displaystyle \int }}^{\text{?}}{e}^x\left[\frac{2+2sinxcosx}{2co{s}^{2}2}\right]dx\left\{\begin{array}{ll}\therefore sin2\theta =2sin\theta cos\theta \hfill & cos2\theta =2co{s}^2\theta -1\hfill \end{array}\right\}$

$={{\displaystyle \int }}^{\text{?}}{e}^x\left[\frac{1}{co{s}^{2}x}+\frac{sinx}{cosx}\right]dx$

$={{\displaystyle \int }}^{\text{?}}{e}^x\left[tanx+se{c}^{2}x\right]dx$ is in the form

${{\displaystyle \int }}^{\text{?}}{e}^x\left[f(x)+f^{\prime }(x)\right]dx$

$f(x)=tanx$

$f^{\prime }(x)=se{c}^{2}x.$

$\therefore I=e^{x}f(x)+c=e^{x}tanx+c.$

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

Let I $={\displaystyle ?f^{?}(ax+b)·[f]{(ax+b)}^n}dx.$

$=\frac{1}{a}{\displaystyle ?[f]{(ax+b)}^n*\left[a{f}^{?}(ax+b)\right]}dx.$

Putting f (ax + b) = t.

$f^{?}(ax+b)\frac{d}{dx}(ax+b)=\frac{dt}{dx}$

af'  (ax + b) dx = dt

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

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

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

Let I$={\displaystyle \int e^{3logx}{\left(x^4+1\right)}^{-1}dx}$

$={\displaystyle \int \frac{e^{log{x}^3}}{\left(x^4+1\right)}dx}$

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

$=\frac{1}{4}log\left|x^4+1\right|+\; C$

Let I = $\int co{s}^{3}x\cdot e^{logsinx}dx.$

$={\displaystyle \int co{s}^{3}xsinxdx}\left\{?e^{logx}=x\right\}$

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

$I\; ={\displaystyle \int t^3(-dt)}=-{\displaystyle \int t^{3}dt}=-\frac{t^4}{4}+C=-\frac{co{s}^{4}x}{4}+C.$

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

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 = -$\frac{1}{3}.$

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

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

$\therefore I=\frac{1}{3}\int \frac{dx}{x^2+1}-\frac{1}{3}\int \frac{dx}{x^2+2^2}$

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

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

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

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

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

$=log|t+1|-log|t+2|+C$

$=log\left|\frac{t+1}{t+2}\right|+C.$

$=log\left[\frac{e^x+1}{e^x+2}\right]+C.$