Answer
$\displaystyle \frac{x^4}{4} \ln x -\frac{x^4}{16}+C$
Work Step by Step
We will solve the given integral by using integrate-by-parts formula such as: $\int udv=uv-\int v du$
$\displaystyle \int x^3 \ln x \ dx=\ln x (\dfrac{x^4}{4})- \int (\frac{x^4}{4} ) \dfrac{1}{x} \ dx$
or, $=\displaystyle \frac{x^4}{4} \ln x -\frac{1}{4} \int x^3 \ dx$
or, $=\displaystyle \frac{x^4}{4} \ln x -\frac{1}{4} (\dfrac{x^4}{4})+C$
or, $=\displaystyle \frac{x^4}{4} \ln x -\frac{x^4}{16}+C$
