Evaluating $\int_{\sqrt{2}}^{\sqrt{5}} \frac{x^3}{\sqrt{x^2-1}} dx$ by substitution 
$$\int_{\sqrt{2}}^{\sqrt{5}} \frac{x^3}{\sqrt{x^2-1}} dx$$

$u^2 = x^2 - 1$
I have worked out that $dx = du$ and that $u = x - 1$ so,
$\int\frac{x^3}{u} du$ - but I'm stuck at this stage.
Any help is appreciated, thanks!

Set $x^2-1 = t$, we then have $2xdx = dt$, which gives us
$$x^3 = \dfrac{x^2 (2xdx)}2 = \dfrac{(1+t)dt}2$$
Hence, the integral becomes
$$\int \dfrac{x^3}{\sqrt{x^2-1}}dx = \int \dfrac{(1+t)dt}{2\sqrt{t}} = \sqrt{t} + \dfrac{t^{3/2}}3 + \text{ constant} = \sqrt{x^2-1} + \dfrac{(x^2-1)^{3/2}}3 + \text{ constant}$$
