$\displaystyle \int 2xln(3x)~dx$

I set:

u=ln(3x)

dv=2x dx

v=x^2

du=1/x

I came up with:

$\displaystyle

x^2ln(3x) - \int x^2 * 1/x~dx

$

But I can't figure out what to do with the part under the integral, because after multiplying I am left with $\displaystyle \int x~dx$. I am sure I messed up somewhere.