Question

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

Answer 1

${\int }_0^{1}ta{n}^{-1}\left(\frac{2x}{1-x^2}\right)dx$

${\int }_0^{1}ta{n}^{-1}\left(tanx\right)dx$

${\int }_0^{1}2xdx=[2\frac{x^2}{2}{]}_0^1=1$