Integrals

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.

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

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

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

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

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

Kindly go through the solution

$I\; ={\displaystyle \int \frac{1}{cos(x+a)cos(x+b)}dx}$

$=\frac{1}{sin(a-b)}\int \frac{sin(a-b)}{cos(x+a)cos(x+b)}·dx$

$=\frac{1}{sin(a-b)}\int \frac{sin[(x+a)-(x+b)]}{cos(x+a)cos(x+b)}dx$

$=\frac{1}{sin(a-b)}·\int \frac{sin(x+a)cos(x+b)-cos(x+a)sin(x+b)}{cos(x+a)cos(x+b)}dx$

$=\frac{1}{sin(a-b)}\left[{\displaystyle \int tan(x+a)dx}-{\displaystyle \int tan}(x+b)dx\right]$

$=\frac{1}{sin(a-b)}[-log|cos(x+a)|+log|cos(x+b)|]+\; C$

$=\frac{1}{sin(a-b)}log\left|\frac{cos(x+b)}{cos(x+a)}\right|+\; C$

$\int \frac{si{n}^{8}x-co{s}^{8}x}{1-2si{n}^{2}xco{s}^{2}x}dx=\int \frac{\left(si{n}^{4}x+co{s}^{4}x\right)\left(si{n}^{4}x-co{s}^{4}x\right)}{\left(si{n}^{2}x+co{s}^{2}x\right)-si{n}^{2}xco{s}^{2}x-si{n}^{2}xco{s}^{2}x}dx$

$=\int \frac{\left(si{n}^{4}x+co{s}^{4}x\right)\left(si{n}^{2}x+co{s}^{2}x\right)\left(si{n}^{2}x-co{s}^{2}x\right)}{si{n}^{2}x\left(1-co{s}^{2}x\right)+co{s}^{2}x\left(1-si{n}^{2}x\right)}dx\{\begin{array}{cc}\hfill ?a^2-b^2=(a+b)(a-b)\hfill & \hfill 1=si{n}^{2}x+co{s}^{2}x.\hfill \end{array}$

$={{\displaystyle \int }}^{\text{?}}\frac{\left(si{n}^{4}x+co{s}^{4}x\right)\left(si{n}^{2}x+co{s}^{4}x\right)(si{n}^{2}x-co{s}^{2}x)*1}{si{n}^{2}x\cdot si{n}^{2}x+co{s}^{2}x\cdot co{s}^{2}x}dx$

$=\int \frac{\left(si{n}^{4}x+co{s}^{2}x\right)\left(si{n}^{2}x-co{s}^{2}x\right)}{\left(sin4x+co{s}^{4}x\right)}dx$

$={\displaystyle \int -}\left(co{s}^{2}x-si{n}^{2}x\right)dx$

$=-{\displaystyle \int cos}2xdx$

$=-\frac{sin2x}{2}+\; C$

Kindly go through the solution

$\int \frac{e^{5logx}-e^{4logx}}{e^{3logx}-e^{2logx}}dx=\int \frac{e^{log{x}^5}-e^{log{x}^4}}{e^{log{x}^3}-e^{log{x}^2}}dx\{\begin{array}{cc}\hfill ?nlogm\hfill & \hfill =log{m}^n\hfill \end{array}$

$=\int \frac{x^5-x^4}{x^3-x^2}dx\{?e^{logx}=x$

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