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

Alternative solution (without trigonometry). Note that by integration by parts
$$\begin{align}
\int \frac{\sqrt{x^2-1}}{x^4}dx&=-\frac{\sqrt{x^2-1}}{3x^3}+\frac{1}{3}\int \frac{1}{x^2\sqrt{x^2-1}}dx\\
&=-\frac{\sqrt{x^2-1}}{3x^3}+\frac{1}{6}\int \frac{D(1-1/x^2)}{\sqrt{1-1/x^2}}dx\\
&=-\frac{\sqrt{x^2-1}}{3x^3}+\frac{\sqrt{x^2-1}}{3x}+c=\frac{(x^2-1)^{3/2}}{3x^3}+c.
\end{align}$$
