Okay, I have a question. Is $\displaystyle \int arctan{5y}$ equal to $\displaystyle \frac{1}{1+5y^2}\times10y$?

I am working on this problem,

$\displaystyle \int x^2arctan{2x}dx$, it's actually tan^-1(2x) but I'm not sure how to insert the negative to make it the inverse.

I made $\displaystyle u=2y,\ du=2dy,\ dv=arctan{5y},\ and\ v=\frac{10y}{a+5y^2}$

Am I heading in the right direction? Thanks in advance.

-Matt