Answer
$$ \frac{1}{3}x^3 \ln x-\frac{1}{9} x^3 +C $$
Work Step by Step
Given $$\int x^{2} \ln x d x$$
Let
\begin{align*}
u&= \ln x\ \ \ \ \ \ dv= x^2 d x\\
du&=\frac{1}{x} dx\ \ \ \ \ \ v=\frac{1}{3}x^3
\end{align*}
Then
\begin{align*}
\int x^{2} \ln x d x&=\frac{1}{3}x^3 \ln x-\frac{1}{3}\int x^2 dx\\
&= \frac{1}{3}x^3 \ln x-\frac{1}{9} x^3 +C
\end{align*}
