Integrals

$\begin{array}{l}{\displaystyle \int x{\left(logx\right)}^2=\cdot {\left(logx\right)}^2{\displaystyle \int xdx}-{\displaystyle \int \frac{d}{dx}{\left(logx\right)}^2\cdot {\displaystyle \int xdxdx}}}\\ ={\left(logx\right)}^2\times \frac{x^2}{2}-{\displaystyle \int 2logx\times \frac{1}{2}\times \frac{x^2}{2}dx}\\ =\frac{x^2}{2}{\left(logx\right)}^2-{\displaystyle \int logx\cdot xdx}\end{array}$

$\begin{array}{l}=\frac{x^2}{2}{\left(logx\right)}^2-\left[logx{\displaystyle \int xdx-{\displaystyle \int \frac{d}{dx}logx{\displaystyle \int xdxdx}}}\right]\\ =\frac{x^2}{2}{\left(logx\right)}^2-\frac{x^2}{2}logx+{\displaystyle \int \frac{x}{2}dx}\\ =\frac{x^2}{2}{\left(logx\right)}^2-\frac{x^2}{2}logx+\frac{x^2}{4}+\; C\end{array}$

$\begin{array}{l}{\displaystyle \int ta{n}^{-1}xdx={\displaystyle \int \left(ta{n}^{-1}x\right)1\cdot dx}}.\\ =ta{n}^{-1}x{\displaystyle \int dx-{\displaystyle \int \frac{d}{dx}ta{n}^{-1}x{\displaystyle \int dxdx}}}\\ =xta{n}^{-1}x-{\displaystyle \int \frac{1}{1+x^2}\cdot xdx}.\\ =xta{n}^{-1}x-\frac{1}{2}{\displaystyle \int \frac{2x}{1+x^2}dx}\\ =xta{n}^{-1}x-\frac{1}{2}\cdot log\left|1+x^2\right|+\; C\end{array}$

$\begin{array}{l}{\displaystyle \int xse{c}^{2}xdx=x{\displaystyle \int se{c}^{2}xdx-{\displaystyle \int \frac{dx}{dx}{\displaystyle \int se{c}^{2}xdx}dx.}}}\\ =xtanx-{\displaystyle \int tanxdx}\\ =xtanx-\left(-log\left|cosx\right|\right)+\; C\\ =xtanx+log\left|cosx\right|+\; C\end{array}$

Kindly go through the solution

$LetI={{\displaystyle \int }}^{\text{?}}{\left(si{n}^{-1}x\right)}^{2}dx$

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

$So,I={{\displaystyle \int }}^{\text{?}}{\theta }^2\cdot cos\theta d\theta ={\theta }^2{{\displaystyle \int }}^{\text{?}}cos\theta d\theta -{{\displaystyle \int }}^{\text{?}}\frac{d}{d\theta }{\theta }^2{{\displaystyle \int }}^{\text{?}}cos\theta d\theta d\theta .$

$={\theta }^{2}sin\theta -{{\displaystyle \int }}^{\text{?}}2\theta sin\theta d\theta .\\ ={\theta }^{2}sin\theta -2{{\displaystyle \int }}^{\text{?}}\theta sin\theta d\theta .$

$\begin{array}{cc}\hfill \hfill & \hfill ={\theta }^{2}sin\theta -2\left[\theta {{\displaystyle \int }}^{\text{?}}sin\theta d\theta -{{\displaystyle \int }}^{\text{?}}\frac{d\theta }{d\theta }{{\displaystyle \int }}^{\text{?}}sin\theta d\theta d\theta \right]\hfill \end{array}\\$

$={\theta }^{2}sin\theta -2\left[\theta (-cos\theta )-{{\displaystyle \int }}^{\text{?}}(-cos\theta )d\theta \right]\\$

$={\theta }^{2}sin\theta +2\theta cos\theta -2sin\theta +C$

$Let,I={\displaystyle \int x}co{s}^{-1}xdx$

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

$So,I={{\displaystyle \int }}^{\text{?}}cos\theta \times \theta \cdot (-sin\theta )d\theta .$

$=-\frac{1}{2}{{\displaystyle \int }}^{\text{?}}\theta (2sin\theta cos\theta )d\theta$ {θsin 2θ = 2 sinθ cosθ}

$\begin{array}{l}=-\frac{1}{2}{{\displaystyle \int }}^{\text{?}}\theta sin2\theta d\theta .\\ =-\frac{1}{2}\left[\theta {{\displaystyle \int }}^{\text{?}}sin2\theta d\theta -{{\displaystyle \int }}^{\text{?}}\frac{d\theta }{d\theta }{{\displaystyle \int }}^{\text{?}}sin2\theta d\theta d\theta \right]\end{array}$

$\begin{array}{l}=-\frac{1}{2}\left[\frac{\theta (-cos2\theta )}{\theta }-{{\displaystyle \int }}^{\text{?}}\left(\frac{(-cos2\theta )}{2}\right)d\theta \right]\\ =\frac{\theta }{4}cos2\theta -\frac{1}{4}\cdot \frac{sin2\theta }{2}+C\\ =\frac{\theta }{4}cos2\theta -\frac{1}{8}(2sin\theta cos\theta )+C.\\ =\frac{\theta }{4}\left[2co{s}^2\theta -1\right]-\frac{1}{4}sin\theta cos\theta +C. \left\{?cos2\theta =2co{s}^2\theta -1\right\}.\\ \end{array}$$\begin{array}{l}\end{array}$

$\begin{array}{l}{\displaystyle \int x}ta{n}^{-1}xdx=ta{n}^{-1}x{\displaystyle \int x}dx-\int \frac{d}{dx}ta{n}^{-1}x·{\displaystyle \int xdx.dx}\\ =ta{n}^{-1}x·\frac{x^2}{2}-\int \frac{1}{1+x^2}·\frac{x^2}{2}dx.\\ =\frac{x^2}{2}ta{n}^{-1}x-\frac{1}{2}\int \frac{x^2}{1+x^2}dx\\ =\frac{x^2}{2}ta{n}^{-1}x-\frac{1}{2}\int \frac{\left(1+x^2\right)-1}{1+x^2}dx\\ =\frac{x^2}{2}·ta{n}^{-1}x-\frac{1}{2}\left[{\displaystyle \int \frac{\left(1+x^2\right)}{1+x^2}}dx-{\displaystyle \int \frac{dx}{1+x^2}}\right]\\ =\frac{x^2}{2}·ta{n}^{-1}x-\frac{1}{2}\left[{\displaystyle \int d}x-ta{n}^{-1}x\right]\\ =\frac{x^2}{2}ta{n}^{-1}x-\frac{1}{2}\left[x-ta{n}^{-1}x\right]+C.\\ =\frac{1}{2}\left[x^{2}ta{n}^{-1}x-x+ta{n}^{-1}x\right]+C.\\ =\frac{1}{2}\left[\left(x^2+1\right)ta{n}^{-1}x·-x\right]+C\end{array}$

Kindly go through the solution

$\begin{array}{l}\int x^2·logx·dx=logx·\int x^{2}dx-\int \frac{d}{dx}logx·\int x^{2}dxdx\\ =logx·\frac{x^3}{3}-\int \frac{1}{x}·\frac{x^3}{3}dx\\ =\frac{x^3}{3}·logx-\int \frac{x^2}{3}dx\\ =\frac{x^3}{3}·logx-\frac{1}{3}\times \frac{x^3}{3}+c\\ =\frac{x^3}{3}logx-\frac{x^3}{9}+c.\end{array}$

$\begin{array}{l}{\displaystyle \int x}log2xdx=log2x·{\displaystyle \int x}dx-\int \frac{d}{dx}log2x{\displaystyle \int x}dxdx\\ =\frac{x^2}{2}·log2x-\int \frac{1}{2x}\times \frac{d(2x)}{dx}\times \frac{x^2}{2}dx+C\\ =\frac{x^2}{2}·log2x-\int \frac{1}{2x}\times 2\times \frac{x^2}{2}{1}^{dx}+C\\ =\frac{x^2}{2}log2x-\frac{1}{2}{\displaystyle \int x}dx+c\\ =\frac{x^2}{2}log2x-\frac{1}{2}·\frac{x^2}{2}+c\\ =\frac{x^2}{2}log2x-\frac{x^2}{4}+c\end{array}$