Answer
$$\frac{1}{39}(z^{3}+1)^{13}+c$$
Work Step by Step
Given $$ \int z^{2}\left(z^{3}+1\right)^{12} d z$$
Let
$$u=z^{3}+1 \ \ \ \Rightarrow \ \ du= 3z^2dz $$
then
\begin{align*}
\int z^{2}\left(z^{3}+1\right)^{12} d z &=\frac{1}{3} \int u^{12} d u \\ &=\frac{1}{3}\left(\frac{1}{13} u^{13}+C\right) \\
&=\frac{1}{39} u^{13}+c\\
&= \frac{1}{39}(z^{3}+1)^{13}+c
\end{align*}
