Question

$\int \frac{x^2{\tan }^{-1}x}{1+x^2}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \left(\frac{x^2{\tan }^{-1}x}{1+x^2}\right)dx \\ =\int \left(\frac{x^2+1-1}{x^2+1}\right){\tan }^{-1}xdx \\ =\int \left(1-\frac{1}{x^2+1}\right){\tan }^{-1}xdx \\ =\int \underset{\mathrm{II}}{1}.\underset{\mathrm{I}}{{\tan }^{-1}}xdx-\int \frac{{\tan }^{-1}x}{x^2+1}dx \\ =\left[{\tan }^{-1}x\int 1dx-\int \left\{\frac{d}{dx}\left({\tan }^{-1}x\right)\int 1dx\right\}dx\right]-\int \frac{{\tan }^{-1}x}{x^2+1}dx \\ =t\left[{\mathrm{an}}^{-1}x\times x-\int \frac{x}{1+x^2}dx\right]-\int \frac{{\tan }^{-1}x}{x^2+1}dx \\ \mathrm{Putting}{x}^2+1=t\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{integral}\mathrm{and}{\tan }^{-1}x=p\mathrm{in}\mathrm{the}\mathrm{second}\mathrm{integral} \\ \Rightarrow 2xdx=dt\mathrm{and}\frac{1}{1+x^2}dx=dp \\ \Rightarrow xdx=\frac{dt}{2}\mathrm{and}\frac{1}{1+x^2}dx=dp \\ \therefore I={\tan }^{-1}x.x-\frac{1}{2}\int \frac{dt}{t}-\int p.dp \\ =x{\tan }^{-1}x-\frac{1}{2}\ln \left|t\right|-\frac{p^2}{2}+C \\ =x{\tan }^{-1}x-\frac{1}{2}\ln \left|1+x^2\right|-\frac{{\left({\tan }^{-1}x\right)}^2}{2}+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\because \mathit{}t=x^{\mathit{2}}+1\mathit{}\mathrm{and}p={\tan }^{-1}x\right] \end{gathered}$$