Answer
$$\frac{1}{6}x^6\ln \left(x\right)-\frac{x^6}{36}+C$$
Work Step by Step
Given
$$\int x^{5} \ln x d x$$
Use the rule
$$\int x^{n} \ln x d x=\frac{x^{n+1}}{(n+1)^{2}}[-1+(n+1) \ln x]+C$$
Here $n=5$, then
\begin{aligned}
\int x^{5} \ln x d x&=\frac{x^{6}}{(6)^{2}}[6 \ln x-1]+C\\
&= \frac{1}{6}x^6\ln \left(x\right)-\frac{x^6}{36}+C
\end{aligned}
