Question

Evaluate the integral of tan^(-1) x dx/(1+x²)

Answer 1

$\int\cfrac{\tan^{-1}(x)dx}{1+x^2} = \int\cfrac{\arctan (x)}{1+x^2}dx = [t = \arctan (x), dt = \cfrac{dx}{1+x^2}] =\\ = \int tdt = \cfrac{t^2}{2}+C|_{t=\arctan(x)}=\cfrac{\arctan^2(x)}{2}+C$