Answer
$\displaystyle \frac{x^3}{3} \ln x -\frac{x^3}{9}+C$
Work Step by Step
We will solve the given integral by using integrate-by-parts formula such as: $\int udv=uv-\int v du$
$\displaystyle \int x^2 \ln x \ dx=\ln x (\dfrac{x^3}{3})- \int (\frac{x^3}{3} ) \dfrac{1}{x} \ dx$
or, $=\displaystyle \frac{x^3}{3} \ln x -\frac{1}{3} \int x^2 \ dx$
or, $=\displaystyle \frac{x^3}{3} \ln x -\frac{1}{3} (\dfrac{x^3}{3})+C$
or, $=\displaystyle \frac{x^3}{3} \ln x -\frac{x^3}{9}+C$
