Answer
$$x(\ln x)^{3}-3 x(\ln x)^{2}+6 x \ln x-6 x+C$$
Work Step by Step
Given
$$ \int(\ln x)^{3} d x$$
by using the form
$$ \int(\ln x)^{n} d x =x(\ln x)^{n}-n \int(\ln x)^{n-1} d x $$
\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 \int(\ln x)^{1} d x\right] \\ &=x(\ln x)^{3}-3 x(\ln x)^{2}+6\left[x(\ln x)^{1}-1 \int(\ln x)^{0} d x\right] \\ &=x(\ln x)^{3}-3 x(\ln x)^{2}+6 x \ln x-6 \int 1 d x\\
&=x(\ln x)^{3}-3 x(\ln x)^{2}+6 x \ln x-6 x+C
\end{align*}
