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..

$\displaystyle\int\frac{\sqrt{x^2-4}}{x^4}dx=\int\frac{\sqrt{1-\frac{4}{x^2}}}{x^3}dx$
Put $\frac1x=z\implies \frac{-1}{x^2}dx=dz$
So it boils to, $-\displaystyle\int z\sqrt{1-4z^2}dz$
Again put $1-4z^2=t\implies -8zdz=dt$
Hence we get, $\frac18\displaystyle\int \sqrt{t}dt=\frac{1}{12}t^{3/2}+C=\frac{1}{12}(1-4z^2)^{3/2}+C=\frac{(1-4x^2)^{3/2}}{12x^3}+C$
