Question

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

Answer 1

`tan^-1x/2 + C`

Explanation:

Let `u = tan^-1x`. Then `du = 1/(x^2 + 1)dx` and `dx = (du)/(1/(x^2 + 1)) = x^2 + 1du`

Substitute:

`=> intu/(1 + x^2) * 1 + x^2 du`

`=> intu du`

Use the power rule `int(x^n) = x^(n + 1)/(n + 1) + C`.

`=> 1/2u^2 + C`

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

`=> tan^-1x/2 + C`

Hopefully this helps!