Answer
$$ \frac{1}{7} (x^{2}-1)^{7 / 2}+\frac{1}{5}(x^{2}-1)^{5 / 2}+C$$
Work Step by Step
Given
$$ \int\left(x^{3}\right)\left(x^{2}-1\right)^{3 / 2} d x $$
Let
$$u=x^{2}-1\ \ \ \Rightarrow \ \ \ du=2xdx $$
Then
\begin{aligned} \int\left(x^{3}\right)\left(x^{2}-1\right)^{3 / 2} d x &=\frac{1}{2} \int(u+1) u^{3 / 2} d u \\ &=\frac{1}{2} \int\left(u^{5 / 2}+u^{3 / 2}\right) d u \\ &=\frac{1}{2}\left(\frac{2}{7} u^{7 / 2}+\frac{2}{5} u^{5 / 2}+C\right) \\ &=\frac{1}{7} u^{7 / 2}+\frac{1}{5} u^{5 / 2}+C\\
&= \frac{1}{7} (x^{2}-1)^{7 / 2}+\frac{1}{5}(x^{2}-1)^{5 / 2}+C \end{aligned}
