# Arctan integration question

∫xarctanx dx

## Homework Equations

d/dx (arctanx) = 1/(1 + x^2)
∫1/(1 + x^2) = arctanx
∫u dv = uv - ∫v du

## The Attempt at a Solution

Hi everyone,

Here's what I've done so far:

Use integration by parts:
u = arctanx
du = 1/(1 + x^2) dx

dv = x dx
v = (1/2)x^2

∫u dv = uv - ∫v du
= arctanx(1/2)x^2 - (1/2)∫x^2/(1 + x^2) dx

But how do you integrate this new integral without going back into arctanx, in which case you'll be left with zero?

Thanks for any help

Long division. $$\frac{x^2}{1+x^2}=1-\frac{1}{1+x^2}$$