Hello, here is the problem I am working on $\displaystyle \int x^2arctan(2x)dx$ I set $\displaystyle u=\ tan^{-1}{2x},\ du=\ \frac {2}{1+4x^2},\ dv=\ x^2,\ and\ v=\ \frac {x^3}{3}$

Now, I got $\displaystyle \frac {x^3tan^{-1}{2x}}{3}-\int \frac {2x^3}{3+12x^2}dx$ How can I integrate $\displaystyle \int \frac {2x^3}{3+12x^2}dx$?

Thanks,

Matt