Evaluate $\int_0^2\frac{x^5}{\sqrt{x^3+6}}\,dx.$ I am stuck on the following integral:
$\displaystyle\int_0^2\dfrac{x^5}{\sqrt{x^3+6}}\,dx.$
I have no idea how one can work it out. Normally I'd try $u=x^3+6$ but this surely does not work here.

$$\frac{x^5}{\sqrt{x^3+6}}=x^3\frac{x^2}{\sqrt{x^3+6}}$$
Set $\sqrt{x^3+6}=u\implies x^3=u^2-6,\dfrac{3x^2dx}{\sqrt{x^3+6}}=du$
