Answer
$$\int x^{n} \ln x d x=\frac{x^{n+1}}{(n+1)^{2}}[-1+(n+1) \ln x]+C$$
Work Step by Step
Given
$$\int x^{n} \ln x d x $$
Use integration by parts , let
\begin{aligned}
u&= \ln x \ \ \ \ \ \ &dv&= x^2 dx\\
du&=\frac{1}{x} dx \ \ \ \ \ \ &v&=\frac{1}{n+1}x^{n+1}dx
\end{aligned}
then
\begin{aligned}
\int x^{n} \ln x d x &= uv-\int vdu \\
&= \frac{1}{n+1}x^{n+1}\ln x-\frac{1}{n+1}\int \frac{x^{n+1}}{x}dx\\
&= \frac{1}{n+1}x^{n+1}\ln x-\frac{1}{n+1}\int x^{n}dx\\
&= \frac{1}{n+1}x^{n+1}\ln x-\frac{1}{n+1}\frac{1}{n+1}x^{n+1}+C\\
&= \frac{x^{n+1}}{(n+1)^{2}}[(n+1)\ln (x)-1 ]+C
\end{aligned}
