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!
 A: 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
$$\int \dfrac{x^3}{\sqrt{x^2-1}}dx = \int \dfrac{(1+t)dt}{2\sqrt{t}} = \sqrt{t} + \dfrac{t^{3/2}}3 + \text{ constant} = \sqrt{x^2-1} + \dfrac{(x^2-1)^{3/2}}3 + \text{ constant}$$
A: Note that, if $u^2=x^2-1$ this does not implies that $du=dx$. Indeed, you have to proceed as follows: $u^2=x^2-1$ gives $ u=\sqrt{x^2-1} $, and so  $du =\frac{xdx}{\sqrt{x^2-1}}$. this implies that your integral become 
$$\int_{1}^{2} u^2+1 du = \frac{8}{3}+2-\frac{1}{3}-1 = \frac{10}{3}$$
A: 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.
