Answer
$$x(\ln x)^{3}-3 x(\ln x)^{2}+6 x \ln x-6 x+C$$
$$x(\ln x)^{2}-2 x(\ln x) +2 x+C$$
Work Step by Step
Since
$$\int(\ln x)^{k} d x=x(\ln x)^{k}-k \int(\ln x)^{k-1} d x$$
Then for $k=2$
\begin{align*}
\int(\ln x)^{2} d x&=x(\ln x)^{2}-2 \int(\ln x) d x\\
&=x(\ln x)^{2}-2\left(x(\ln x) - \int d x \right)\\
&=x(\ln x)^{2}-2 x(\ln x) +2 x+C \\
\end{align*}
For $k=3$
\begin{align*}
\int(\ln x)^{3} d x&=x(\ln x)^{3}-3 \int(\ln x)^2 d x\\
&=x(\ln x)^{3}-3\left( x(\ln x)^{2}-2 x(\ln x) +2 x \right)\\
&= x(\ln x)^{3}-3 x(\ln x)^{2}+6 x \ln x-6 x+C\\
\end{align*}
