# 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.

Let $u = x^3 + 6 \implies du = 3x^2 \,dx$, and $x^3 = u - 6$

That gives us $$\frac 13\int\dfrac{3x^2(x^3)\,dx}{\sqrt {x^3 + 6}} = \frac 13\int{(u-6)}u^{-1/2}\,du$$

Can you take it from here?

$\text{hint}$

$$\frac{x^5}{\sqrt{x^3+6}} = \frac{x^3\cdot x^2}{\sqrt{x^3+6 }}$$

$$\frac{x^5}{\sqrt{x^3+6}}=x^3\frac{x^2}{\sqrt{x^3+6}}$$

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

Let $u^2 = x^3 + 6$.... so that the denominator will eventually simplify as follows

$x^6 = (u^2-6)^2$

Differentiate $6 x^5 dx = 2 (u^2-6) \,2 u\, du$

Now, integral simplifies much to: $\frac 23\int (u^2-6) du = \frac 23 (u^3/3 - 6\, u )$

Limits of x: $(0,2)$ , Limits of u: $(\sqrt 6, \sqrt 14)$. And now, your turn :)

$\frac 89 (3 \sqrt 6 - \sqrt 14)$