Integrals

Kindly go through the solution

$\begin{array}{l}LetI={\displaystyle \int x^2{e^x}^{3}dx.}\\ let,t=x^3\Rightarrow dt=3x^{2}dx\cdot \Rightarrow \frac{dt}{3}=x^{2}dx\\ So,={\displaystyle \int \cdot }{e}^t\frac{dt}{3}=\frac{1}{3}{\displaystyle \int e^t}dt.\\ =\frac{1}{3}{e}^t+\; C\\ =\frac{e^{x^3}}{3}+C.\end{array}$

So, option (A) is correct.

$Let,\; I={\displaystyle \int si{n}^{-1}}\left(\frac{2x}{1+x^2}\right)dx$

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

$I={\displaystyle \int si{n}^{-1}}\left[\frac{2tan\theta }{1+ta{n}^2\theta }\right]\cdot se{c}^2\theta d\theta$

$\begin{array}{l}Let,\; I={\displaystyle \int e^{2x}}sinxdx.\\ =sinx{\displaystyle \int e^{2x}}-{\displaystyle \int \frac{d}{dx}}sinx{\displaystyle \int e^{2x}}dxdx\\ =sinx\cdot \frac{e^{2x}}{2}-{\displaystyle \int cos}x\cdot \frac{e^{2x}}{2}dx\\ =\frac{e^{2x}}{2}sinx-\frac{1}{2}{\displaystyle \int cosx}\cdot *e^{2x}dx\\ =\frac{e^{2x}}{2}\cdot sinx-\frac{1}{2}\left\{cosx{\displaystyle \int e^{2x}}dx-{\displaystyle \int \frac{d}{dx}}cosx{\displaystyle \int e^{2x}}dxdx\right\}\\ =\frac{e^{2x}}{2}sinx-\frac{1}{2}\left\{cosx\cdot \frac{e^{2x}}{2}+{\displaystyle \int sinx\frac{e^{2x}}{2}dx}\right\}\\ =\frac{e^{2x}}{2}sinx-\frac{e^{2x}}{4}cosx-\frac{1}{4}I+\; C\\ \Rightarrow I+\frac{I}{4}=\frac{e^{2x}}{2}sinx-\frac{e^{2x}}{4}cosx+\; C\\ \Rightarrow \frac{5I}{4}=\frac{e^{2x}}{2}sinx-\frac{e^{2x}}{4}cosx+\; C\\ \Rightarrow I=\frac{4}{5}\left[\frac{e^{2x}}{2}sinx-\frac{e^{2x}}{4}cosx\right]+\; C\\ =\frac{e^{2x}}{5}[2sinx-cosx]+\; C\end{array}$

$\begin{array}{l}Let,\; I\; ={\displaystyle \int \frac{x-3}{{(x-1)}^3}}{e}^{x}dx.\\ ={\displaystyle \int \frac{(x-1)-2}{{(x-1)}^3}}{e}^{x}dx.\\ ={\displaystyle \int e^x}\left[\frac{x-1}{{(x-1)}^3}-\frac{2}{{(x-1)}^3}\right]dx\\ ={\displaystyle \int e^x}\left[\frac{1}{{(x-1)}^2}+\frac{(-2)}{{(x-1)}^3}\right]dx,whichisinfore\\ {\displaystyle \int e^x}\left[f(x)+f^{\prime }(x)\right]dx,Where\\ f(x)=\frac{1}{{(x-1)}^2}\\ f^{\prime }(x)=\frac{d}{dx}{(x-1)}^{-2}=\frac{-2}{{(x-1)}^3}.\\ \therefore I=e^{x}f(x)+c=e^x\frac{1}{{(x-1)}^2}+c=\frac{e^x}{{(x-1)}^2}+\; C\end{array}$

$\begin{array}{l}Let,\; I={\displaystyle \int e^x}\left(\frac{1}{x}-\frac{1}{x^2}\right)dx{\displaystyle \int e^x}[f(x)+f(x)]dx\\ Where,f(x)=\frac{1}{x}\\ f(x)=\frac{dx}{dx}=-1x^{-2}=\frac{-1}{x^2}\end{array}$

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

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

$Let,={\displaystyle \int e^x}\left(\frac{1+sinx}{1+cosx}\right)dx.,$

$={\displaystyle \int e^x}\frac{1+2sin\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.cos\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{2co{s}^2\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}dx.$ {∴ sin 2x = 2sin x cos x. cos 2x = cos^{- 2}x- 1 1 + cos 2x = 2cos²x.

$\begin{array}{l}={\displaystyle \int e^x}\left[\frac{1}{2co{s}^2\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{2sin\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.cos\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{2co{s}^2\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]dx]\\ ={\displaystyle \int e^x}\left[\frac{1}{2}se{c}^2\frac{x}{2}+\frac{cosx\cdot sin\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{cos\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]dx\end{array}$

$={\displaystyle \int e^x}\left[tan\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+\frac{1}{2}si{n}^2\frac{x}{2}\right]dx$ is in the form

  e^x [f(x) + f(x)].dx where

$\begin{array}{l}f(x)=tan\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ f^{\prime }(x)=\frac{d}{dr}tan\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=se{c}^2\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\frac{d}{dx}\left(\frac{x}{2}\right)\\ =\frac{1}{2}se{c}^2\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ =e^{x}f(x)+c=e^{x}tan\frac{x}{2}+\; C\end{array}$

$Let,I={\displaystyle \int \frac{x{e}^x}{{(x+1)}^2}}dx={\displaystyle \int \frac{x+1-1}{{(x+1)}^2}}{e}^{x}dx$

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

e^x [f(x) + f(x)] dx

$\begin{array}{l}Where,f(x)=\frac{x+1}{{(x+1)}^2}=\frac{1}{x+1}\\ So,f^{\prime }(x)=\frac{d{(x+1)}^{-1}}{dx}=\frac{(-1)}{{(x+1)}^2}.\\ \therefore {\displaystyle \int \frac{x^{\cdot }\cdot e^x}{\left(1+x^2\right)}}dx=e^x\left[\frac{1}{x+1}\right]+\; C\end{array}$

∴f (x) = sin x

f (x) = cos x.

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

${\displaystyle \int e^x(sinx+cosx)dx=e^x}sinx+c$

$\begin{array}{l}{\displaystyle \int \left(x^2+1\right)}logxdx=logx{\displaystyle \int \left(x^2+1\right)dx-{\displaystyle \int \left(x^2+1\right)dx-{\displaystyle \int \frac{d}{dx}logx{\displaystyle \int \left(x^2+1\right)dxdx}}}}\\ =logx\cdot \left[\frac{x^3}{3}+x\right]-{\displaystyle \int \frac{1}{x}*\left[\frac{x^3}{3}+x\right]dx}\\ =\left[\frac{x^3}{3}+x\right]logx-{\displaystyle \int \left[\frac{x^2}{3}+1\right]dx}\\ =\left[\frac{x^3}{3}+x\right]logx-\frac{x^3}{3*3}-x+\; C\\ =\left[\frac{x^3}{3}+x\right]logx-\frac{x^3}{9}-x+\; C\end{array}$