Question

How do you integrate `int tan^-1x/(x^2+1)` by trigonometric substitution?

Answer 1

The answer is `=(arctanx)^2/2+C`

Explanation:

Apply the substitution

`u=arctanx`, `=>`, `du=(dx)/(1+x^2)`

Therefore,

`int(arctanxdx)/(1+x^2)=int(u*(1+x^2)du)/(1+x^2)=intudu`

`=u^2/2`

`=(arctanx)^2/2+C`