Question

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

Answer 1

As we know that,

$2{\tan }^{-1}x={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$

Therefore,

$\int {\tan }^{-1}\left(\frac{2x}{1-x^2}\right)dx=2\int {\tan }^{-1}xdx$

Now,

$\int {\tan }^{-1}xdx$

$=\int (1\times {\tan }^{-1}x)dx$

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

$=x\cdot {\tan }^{-1}x-\frac{1}{2}\ln (x^2+1)+c(\because \frac{d}{dx}(x^2+1)=2x)$

Therefore,

$\int {\tan }^{-1}\left(\frac{2x}{1-x^2}\right)dx=2x\cdot {\tan }^{-1}x-\ln (x^2+1)+c$