Answer
$$ \frac{x^4}{16}(4\ln x -1).$$
Work Step by Step
We have $u=\ln x$ and $dv=x^3dx$. Then, $du=\frac{dx}{x}$, $v=\frac{x^4}{4}$
$$ \int x^3 \ln x dx=\int udv=uv-\int vdu\\
= \frac{x^4\ln x}{4}- \int \frac{x^3}{4} dx\\
=\frac{x^4\ln x}{4}- \frac{x^4}{16} \\
= \frac{x^4}{16}(4\ln x -1).$$
