Integrate $\int \frac{\sqrt{x^2-1}}{x^4}dx$ I am trying to integrate $\int \frac{\sqrt{x^2-1}}{x^4}dx$ via trig substitution. I decided to substitute $x = \sec\theta$ into the square root and $dx = \sec\theta \tan\theta\,d\theta$.
$$\int \frac{\sqrt{\sec^2 \theta-1^2}}{\sec^4\theta} \,dx
= \int \frac{\sqrt{\tan^2\theta + 1 - 1}}{\sec^4\theta}\,dx 
= \int \frac{\tan\theta}{\sec^4\theta} \sec\theta \tan\theta\,d\theta = \int \dfrac{\tan^2\theta}{\sec^3\theta}\,d\theta$$
Here is where I am currently stuck. I attempted substitution with $u = \sec\theta, du = \sec x \tan x dx$ but that didn't seem to work out. I wasn't able to get an integration by parts strategy working either.
I think the answer lies in some sort of trigonometry regarding $\int \frac{\tan^2\theta}{\sec^3\theta}\,d\theta$ that I am overlooking to further simplify the problem, but no idea what it is

$$
\begin{aligned}
& \int \frac{\sqrt{x^{2}-1}}{x^{4}} d x \\\stackrel{y=\frac{1}{x}}{=}  &\int\frac{\sqrt{\frac{1}{y^{2}}-1}}{\frac{1}{y^{4}}}\left(-\frac{1}{y^{2}} d y\right)\\
=& -\int y \sqrt{1-y^{2}}d y \\
=&\frac{\left(1-y^{2}\right)^{\frac{3}{2}}}{3}+C
\\=&\frac{\left(x^{2}-1\right)^{\frac{3}{2}}}{3 x^{3}}+C
\end{aligned}
$$
