Answer
$$\frac{1}{3}\left( (x^3-1)+\ln{|x^3-1|} \right)+C$$
Work Step by Step
Given $$\int \frac{x^{5}}{x^{3}-1} d x$$
Let $$u=x^3-1\ \ \ \ \ \ \ \ du=3x^2dx $$Then
\begin{align*}
\int \frac{x^{5}}{x^{3}-1} d x&=\frac{1}{3}\int \frac{u+1}{u} du\\
&=\frac{1}{3}\int\left( 1+\frac{1}{u} \right)du\\
&=\frac{1}{3}\left( u+\ln{u} \right)+C\\
&=\frac{1}{3}\left( (x^3-1)+\ln{|x^3-1|} \right)+C
\end{align*}
