Can someone please help me with this problem I am stuck! (Headbang)
$\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!!!!! (Rock)