$x = a\sec\theta, dx = \sec\theta \tan\theta$
$\int \frac{dx}{(x^2-1)^{\frac{3}{2}}}$ =
$ \int \frac{dx}{(\tan^2\theta)^{\frac{3}{2}}}$ =
$ \int \frac{dx}{\tan^{\frac{7}{2}}\theta}$ =
$\int \frac{\sec\theta \tan\theta}{\tan\theta ^{\frac{7}{2}}}$ =
$\int \tan^{\frac{-5}{2}}\theta \sec\theta$
Here is where I get stuck…I tried converting $\tan\theta$ and $\sec\theta$ in terms of $\cos\theta$ and $\sin\theta$, but that didn't seem to get me anywhere…What is my next move from here? Did I even start this problem correctly? I can't tell 🙁
Update with more work after initial answers:
$\int \frac{\cos\theta}{\sin^2\theta}$ $u = \sin\theta, du = \cos\theta d\theta$
I found $\sin^{-1}\theta = \frac{\sqrt{x^2-1}}{x}$
$= \int \frac{du}{u^2} = \frac{1}{ \frac{1}{3}u^3} =
\frac{1}{3\sin^3\theta}
= 3 \bigg( \frac{x}{\sqrt{x^2-1}} \bigg)^3$
Best Answer
$$\int \frac{dx}{(x^2-1)^{\frac{3}{2}}} = \int \frac{\tan\theta\sec\theta d\theta}{\underbrace{(\tan^2\theta)^{\frac{3}{2}}}_{\tan^3\theta}} = \int \frac{\sec\theta d\theta}{\tan^2\theta}= -\frac1{\sin \theta}=-\frac{x}{\sqrt{x^2-1}}.$$