<s>[INST] 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!
 [/INST] 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 
$$\int_{1}^{2} u^2+1 du = \frac{8}{3}+2-\frac{1}{3}-1 = \frac{10}{3}$$
</s>