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..

Besides to @Ron's answer, you can see that the form Differential binomial is ruling the integral here. Let to write integrand as follows:
$$\int(x^2-4)^{1/2}x^{-4}~dx$$
So, $m=-4,~~p=1/2,~~n=2$ and so $\frac{m+1}{n}+p=-1\in\mathbb Z$ and so the method says that you can use the following nice substitution:
$$x^2-4=t^{2}x^4$$
