Evaluate the integral. $\int x^2 \log(4x) dx$ The problem is $\int x^2 \log(4x) dx$ 
Here $\ln$ refers to the natural logarithm.
So far, I know $u = x^2$ and $du = 2x (dx)$.
So $dv = \ln(4x) dx$ and $v = 1/x$, but I don't know where to go from here. 

Integrating by parts,
$$\int x^2\log 4xdx = \frac{1}{3}x^3\log4x-\int\frac{1}{3}x^3\cdot\frac{4}{4x}dx\\
=\frac{1}{3}x^3\log4x - \frac{1}{3}\int x^2dx\\
=\frac{1}{3}x^3\log4x - \frac{1}{9}x^3 + C$$
