$\displaystyle

\int \frac{\sqrt{x^{2}-4}}{x^2}dx

$

I started with setting

$\displaystyle x=2sec \theta $ and $\displaystyle dx=2sec \theta tan \theta $

and $\displaystyle x^{2} = 4sec^{2} \theta$

$\displaystyle \tan^{2} \theta = \sec^{2} \theta -1 $

Then I rewrite it...

$\displaystyle

\int \frac{2tan\theta}{4sec^{2}\theta} * 2 sec \theta tan\theta d\theta = \int \frac{tan^{2}\theta}{sec\theta}d\theta = \int \frac{sin^{2}\theta}{cos\theta}d\theta

$

When I go on I do not get anything what helps me solving it.

Thanks for any help...