<s>[INST] How do I integrate this expression? Need to find the following: $$\int \frac{\sqrt{x^2 - 4}}{x^3} dx$$
So far, after using integration by parts, I've arrived at $\frac{-\sqrt{x^2-4}}{2x^2} + \frac12 \int \frac{1}{x\sqrt{x^2-4}} dx$. How to proceed from here?
 [/INST] $$I=\int\dfrac{\sqrt{x^2-4}}{x^3}dx=\int\dfrac{\sqrt{x^2-4}}{x^4}x\ dx$$
Set $\sqrt{x^2-4}=u\implies x^2-4=u^2\iff x\ dx=u\ du$
$$I=\int\dfrac{u^2}{(u^2+4)^2} du=\int\dfrac1{(u^2+4)} du-4\int\dfrac{du}{(u^2+4)^2}$$
For the second part set $u=2\tan\theta$

Alternatively, set  $x=2\sec\theta$
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