Integration $\int \frac{\sqrt{x^2-4}}{x^4}$ Problem : 
Integrate $\int \frac{\sqrt{x^2-4}}{x^4}$
I tried : Let $x^2-4 =t^2 \Rightarrow 2xdx = 2tdt$
$\int \frac{\sqrt{x^2-4}}{x^4} \Rightarrow \frac{t^3 dt}{\sqrt{t^2+4}(t^4-8t+16)}$
But I think this made the integral too complicated... please suggest how to proceed.. Thanks..

Use a hyperbolic cosine substitution: $x=2 \cosh{t}$; $dx = 2 \sinh{t}\, dt$.  Then the integral is equal to
$$\frac14 \int dt \frac{\sinh^2{t}}{\cosh^4{t}} = \frac14 \int dt \, \left (\text{sech}^2{t} -  \text{sech}^4{t}\right ) $$
This integral is easy once you recognize that $d(\tanh{t}) = \text{sech}^2{t}$ and $1-\text{sech}^2{t}=\tanh^2{t}$:
$$\frac14 \int dt \, \left (\text{sech}^2{t} -  \text{sech}^4{t}\right ) = \frac14 \int d(\tanh{t}) \tanh^2{t} = \frac{1}{12} \tanh^3{t}+C  $$
where $C$ is a constant of integration.  Then just substitute back to get the integral in $x$.
