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..
 A: 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:
$$x^2-4=t^{2}x^4$$
A: 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 $x$.
A: $\displaystyle\int\frac{\sqrt{x^2-4}}{x^4}dx=\int\frac{\sqrt{1-\frac{4}{x^2}}}{x^3}dx$
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, $\frac18\displaystyle\int \sqrt{t}dt=\frac{1}{12}t^{3/2}+C=\frac{1}{12}(1-4z^2)^{3/2}+C=\frac{(1-4x^2)^{3/2}}{12x^3}+C$
A: HINT:
As the radical contains $x^2-4,$
put $x=2\sec\theta$
$$\implies \int\frac{\sqrt{x^2-4}}{x^4}dx=\int \frac{2|\tan\theta|}{\sec^4\theta} 2\sec\theta\tan\theta d\theta=4\cdot \text{sign}(\tan\theta)\int \sin^2\theta \cos \theta d\theta$$
Putting $\sin\theta=u,$
$$4\text{sign}(\tan\theta)\int \sin^2\theta \cos \theta d\theta$$
$$=4\text{sign}(\tan\theta)\int u^2 du=\frac43\cdot\text{sign}(\tan\theta)\cdot u^3+K$$
$$=\frac43\cdot\text{sign}(\tan\theta)\cdot \sin^3\theta+K$$ where $K$ is an arbitrary  constant of indefinite integral
Now, $\sin\theta=\pm\sqrt{1-\cos^2\theta}=\pm\sqrt{1-\left(\frac2x\right)^2}=\pm\frac{\sqrt{4-x^2}}x$
Now observe that the sign$(\sin\theta)=$ sign$(\tan\theta)\cdot$sign$\left(\frac2x\right)$
