Question

What is `int tan^-1 x dx`?

Answer 1

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

Explanation:

We want to solve

`I=inttan^-1(x)dx`

Use integration by parts / partial integration

`intudv=uv-intvdu`

Let `u=tan^-1(x)` and `dv=1dx`

Then `du=1/(x^2+1)dx` and `v=x`

`I=tan^-1(x)x-intx/(x^2+1)dx`

Make a substitution `u=x^2+1=>(du)/dx=2x`

`I=tan^-1(x)x-1/2int1/(u)du`

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

Substitute back `u=x^2+1`

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