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!

There is another step using integration by part
$$\int \frac{x^3}{\sqrt{x^2-1}}dx = \int x^2\frac{x}{\sqrt{x^2-1}}dx$$
You want to integrate $x(x^2-1)^{-1/2}$ (and differentiate $x^2$), consider product rule. (Substitution is using the idea of product rule, i.e. you can let $u=x^2-1$)
$$\int x(x^2-1)^{-1/2} = (x^2 - 1)^{1/2}dx$$
So the initial integration become
$$\int \frac{x^3}{\sqrt{x^2-1}}dx = x^2(x^2 - 1)^{1/2} - \int x(x^2-1)^{1/2}dx$$
Do integration by part again to yield final answer.
