TITLE: $\int \frac{dx}{x^3+x^2\sqrt{x^2-1}-x}$
QUESTION [5 upvotes]: solve:
$$\int \frac{\mathrm dx}{x^3+x^2\sqrt{x^2-1}-x}$$
I tried:
$$\begin{align}\int \frac{\mathrm dx}{x(x^2-1)+x^2\sqrt{x^2-1}}&=\int \frac{\mathrm dx}{x(\sqrt{x^2-1}\sqrt{x^2-1})+x^2\sqrt{x^2-1}}\\&=\int \frac{\mathrm dx}{x\sqrt{x^2-1}(\sqrt{x^2-1}+x)}\end{align}$$
$x=\sin t$
$$\int \frac{\mathrm dt}{\cos t\sin t-\sin^2t}$$
And I can not continue from here.

REPLY [3 votes]: Hint: Put $x＝\dfrac{1}{\cos t}$, then 
$$\int \frac{\cos^3 t}{\sin t(\sin t+1)}\frac{\sin t}{\cos^2 t}dt＝\int \frac{d\sin t}{\sin t+1}$$