Question

$\int \frac{\left(x{\tan }^{-1}x\right)}{{\left(1+x^2\right)}^{3/2}}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{x{\tan }^{-1}x}{{\left(1+x^2\right)}^{{\displaystyle \frac{3}{2}}}}dx \\ \mathrm{Putting}x=\tan \theta \\ \Rightarrow dx={\sec }^2\theta d\theta \\ \&\theta ={\tan }^{-1}x \\ \therefore I=\int \frac{\left(\tan \mathit{}\theta \right).\theta .{\sec }^{\mathit{2}}\mathit{}\theta d\theta }{{\left(1+{\tan }^2\theta \right)}^{{\displaystyle \frac{3}{2}}}} \\ =\int \frac{\theta .\tan \mathit{}\theta {\sec }^2\theta \mathit{}d\theta }{{\left({\sec }^2\theta \right)}^{{\displaystyle \frac{3}{2}}}} \\ =\int \frac{\theta \mathit{}\tan \mathit{}\theta .{\sec }^2\theta \mathit{}d\theta }{{\sec }^3\theta } \\ =\int \frac{\theta .\tan \theta }{\sec \theta }d\theta \\ =\int \mathit{}\underset{\mathit{I}}{\mathit{\theta }}.\underset{\mathrm{II}}{\sin \theta }d\theta \\ =\theta \int \sin \theta d\theta -\int \left\{\frac{d}{d\theta }\left(\theta \right)\int \sin d\theta \right\}d\theta \\ =\theta \left(-\cos \theta \right)-\int 1.\left(-\cos \mathit{}\theta \right)\mathit{}d\theta \\ =-\theta \cos \theta +\sin \theta +C \\ =\frac{-\theta }{\sec \theta }+\frac{1}{\mathrm{cose}c\mathit{}\theta }+C \\ =\frac{-\theta }{\sqrt{1+{\tan }^2\theta }}+\frac{1}{\sqrt{1+{\cot }^2\mathit{}\theta }}+C \\ =\frac{-\theta }{\sqrt{1+{\tan }^2\theta }}+\frac{\tan \theta }{\sqrt{{\tan }^2\theta +1}}+C \\ =\frac{-{\tan }^{-1}x}{\sqrt{1+x^2}}+\frac{x}{\sqrt{x^2+1}}+C \end{gathered}$$