Question

What's the integral of `int arctan(x) dx`?

Answer 1

`xarctanx-ln(x^2+1)/2+C`

Explanation:

Problem:`intarctanx`
Integrate by parts: `intfgprime=fg-intfprimeg`
`f=arctanx,gprime=1`
`darr`
`fprime=1/(x^2+1),g=x:`
=`xarctanx-intx/(x^2+1)dx`

Now solving:
`intx/(x^2+1)dx`
Substitute `u=x^2+1->dx=1/(2x)du`
`=1/2int1/u`du

Now solving:
`int1/u du`
This is a standard integral
=`lnu`

Plug in solved integrals:
`1/2int1/udu`
=`lnu/2`

Undo substitution `u=x^2+1`:
=`ln(x^2+1)/2`

Plug in solved integrals:
=`xarctanx-intx/(x^2+1)dx`
=`xarctanx-ln(x^2+1)/2`

The problem is solved:
`intarctanx`
=`xarctanx-ln(x^2+1)/2+C`

Answer 2

`int arctan(x) dx = xarctan(x)-1/2log(1+x^2)+C`

Explanation:

Integration by parts with `u=arctan(x)` and `(dv)/(d x)=1`, giving `(du)/(d x) = 1/(1+x^2)` and `v=x`.

Thus
`int arctan(x) dx = xarctan(x)-int x/(1+x^2) dx`

The second term integrates easily as a natural logarithm:
`int arctan(x) dx = xarctan(x)-1/2log(1+x^2)+C`