$\displaystyle u' = \frac{1}{x}$
$\displaystyle v = \frac{46}{3}x^3$
$\displaystyle \int u\,dv = uv - \int v\,du$
Your integral is therefore equal to
$\displaystyle \frac{46}{3}x^3 \cdot \ln(x) - \int \left(\frac{46}{3}x^3 \cdot \frac{1}{x}\right)\,dx$
Use algebra to simplify the term inside the brackets, then evaluate the integral which is a standard power rule one