Answer
$\frac{1}{8} (x^2-1)^4$
Work Step by Step
Let $u=x^2-1$. Then $du=2x\ dx, \Rightarrow \frac{1}{2} \ du=x\ dx $. Now we can replace the integral with:
$$\frac{1}{2} \int u^3 \ du$$
$$=\frac{1}{2} (\frac{1}{4}u^4)$$
$$=\frac{1}{8}u^4$$
$$=\frac{1}{8}(x^2-1)^4$$
(Since C is taken to be 0, we can ignore it.)