I'm having some trouble with a more complicated problem.

$\displaystyle \int \frac {x^{3x}} {\sqrt {4 - e^{6x}}}$

So far I have:

$\displaystyle u = e ^{3x}$ so $\displaystyle \frac {1}{3}du = e^{3x}dx$

That gives me:

$\displaystyle \frac {1}{3}\int \frac {x^{3x}}{\sqrt{4-u^2}}dx$

I need help getting rid of that x on top and turn that 4 into a 1. I'm sure that I'm missing something obvious.