Question

What is the Integral of `tan^-1x/(1+x^2)`?

Answer 1

`= 1/2 (tan^-1 x )^2 + C`

Explanation:

`int \ tan^-1x/(1+x^2) \ dx`

`= int \ tan^-1 x * d/dx (tan^(-1) x) \ dx`

`= int d/dx (1/2(\ tan^-1 x )^2 ) \ dx`

`= 1/2 (tan^-1 x )^2 + C`