I used integration by parts, with $$u = x, du = dx$$ $$dv = \sqrt{1-x^3}dx, v = ??$$
Then to find v, I would need to integrate $\sqrt{1-x^3}dx$. I immediately thought to use trig substitution, but with the $x^3$, it didn't work out.
How would I continue this method if it is correct, or is there another way to approach solving this question?
On
This has no elementary antiderivative. It can be written as an Incomplete Beta Function: $$ \begin{align} \int_0^xt\sqrt{1-t^3}\,\mathrm{d}t &=\frac13\int_0^{x^3}t^{-1/3}\left(1-t\right)^{1/2}\,\mathrm{d}t\\[6pt] &=\frac13\operatorname{B}\left(x^3;\frac23,\frac32\right) \end{align} $$ If $x=1$, we can write this as a ratio of Gamma functions: $$ \begin{align} \frac13\operatorname{B}\left(1;\frac23,\frac32\right) &=\frac13\operatorname{B}\left(\frac23,\frac32\right)\\ &=\frac13\frac{\Gamma\left(\frac23\right)\Gamma\left(\frac32\right)}{\Gamma\left(\frac{13}6\right)}\\ &=\frac{6\sqrt\pi}7\frac{\Gamma\left(\frac23\right)}{\Gamma\left(\frac16\right)} \end{align} $$
As said in comments, this is a very difficult integral, the solution of which involving elliptic integrals (which are not the most funny things to handle).
One solution could be to expand the integrand as a Taylor series (or use the general binomial theorem) using $$\sqrt{1+y} =(1+y)^{1/2} = \sum_{k=0}^{\infty} \binom{1/2}{k}y^k$$ Replacing $y$ by $-x^3$, this gives $$\sqrt{1-x^3}=\sum_{k=0}^{\infty} \binom{1/2}{k}(-1)^k x^{3k}$$ So $$x\sqrt{1-x^3}=\sum_{k=0}^{\infty} \binom{1/2}{k}(-1)^k x^{3k+1}$$ and integrating $$\int x\sqrt{1-x^3}\,dx=\sum_{k=0}^{\infty} \binom{1/2}{k}(-1)^k \int x^{3k+1}\,dx=\sum_{k=0}^{\infty} \binom{1/2}{k}(-1)^k \frac {x^{3k+2}}{3k+2}$$
Edit
As robjohn commented, it is worth to remember that $$\binom{1/2}{k}=\frac{(-1)^{k-1}}{4^k(2k-1)}\binom{2k}{k}$$ All of this makes $$\int x\sqrt{1-x^3}\,dx=-\sum_{k=0}^{\infty} \frac{ \binom{2 k}{k} }{4^k(2 k-1) (3 k+2)}x^{3 k+2}$$ We could also write $$\binom{1/2}{k}=\frac{\sqrt{\pi }}{2 \,\Gamma \left(\frac{3}{2}-k\right)\,k!}$$