# Math Help - Another Integral w/ integration by parts

1. ## Another Integral w/ integration by parts

Hello, here is the problem I am working on $\int x^2arctan(2x)dx$ I set $u=\ tan^{-1}{2x},\ du=\ \frac {2}{1+4x^2},\ dv=\ x^2,\ and\ v=\ \frac {x^3}{3}$
Now, I got $\frac {x^3tan^{-1}{2x}}{3}-\int \frac {2x^3}{3+12x^2}dx$ How can I integrate $\int \frac {2x^3}{3+12x^2}dx$?
Thanks,
Matt

2. Assuming that your work is okay, then just put $z=3+12x^2.$

3. I did this (minus the constants):

In another post I found the "adding 0" method helpful. I tried it with this one and it gave me another answer. I'm not sure if they're the same but are written in different forms though:

4. Use long division and get

$\frac{x}{6} - \frac{x}{24x^2+6}$

You should be able to finish up from here