Can someone please help me with this problem I am stuck!

$\displaystyle

\int 3arctan(2x)dx

$

My first step has been:

dv = 3 dx

v= 3x

u = arctan(2x)

du = [1/(1+4x^2)]*2

Giving me...

$\displaystyle

3xarctan(2x) - \int 3x*[1/(1+4x^2)]*2dx$

Thanks!!!!!