Question

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

Answer 1

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

Let $x=\tan \theta \Rightarrow \sec \theta =\sqrt{1+{\tan }^2\theta }=\sqrt{1+x^2}$

$\therefore \int {\tan }^{-1}\left(\frac{2\tan \theta }{1-\tan \theta }\right).{\sec }^2\theta d\theta$

$=\int {\tan }^{-1}\left(\tan 2\theta \right).{\sec }^2\theta d\theta$

$=\int 2\theta {\sec }^2\theta d\theta$

$=2\theta \tan \theta -2\int \tan \theta d\theta$

$=2\theta \tan \theta -2\ln |\sec \theta |+c$

$=2x{\tan }^{-1}x-2\ln |\sqrt{1+x^2}|+c$