Question

$\underset{0}{\overset{1}{\int }}{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)dx$

Answer 1

$$\begin{gathered} {\int }_0^1{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)dx \\ ={\int }_0^{1}2{\tan }^{-1}x \\ =2{\left[x{\tan }^{-1}x\right]}_0^1-2{\int }_0^1\frac{x}{1+x^2}dx \\ =2{\left[x{\tan }^{-1}x\right]}_0^1-{\left[\log \left(1+x^2\right)\right]}_0^1 \\ =2\frac{\mathrm{\pi }}{4}-0-\log 2+0 \\ =\frac{\mathrm{\pi }}{2}-\log 2 \end{gathered}$$