I am trying to find $\int \frac {\sqrt {x^2 - 4}}{x} dx$
I make $x = 2 \sec\theta$
$\int \frac {\sqrt {4(\sec^2 \theta - 1)}}{x} dx$
$\int \frac {\sqrt {4\tan^2 \theta}}{x} dx$
$\int \frac {2\tan \theta}{x} dx$
From here I am not too sure what to do but I know I shouldn't have x.
$\int \frac {2\tan \theta}{2 \sec\theta} dx$
I also know I shouldn't have dx anymore.
$dx = 2\sec \theta \tan \theta \; \mathrm d\theta$
$\int \frac {2\tan \theta}{2 \sec\theta} 2\sec \theta \tan \theta \; \mathrm d\theta$
$\int {2\tan^2 \theta} \; \mathrm d\theta$
$2\int {\tan^2 \theta} \; \mathrm d\theta$
I have no idea how to find the integral of $\tan^2 \theta$
So I use Wolfram Alpha:
$\tan \theta - \theta + c$
Now I need to replace theta with x.
$x = 2 \sec\theta$
With same mathmagics I produce
$ \frac {x}{2} = \sec \theta$
$ \theta = \operatorname {arcsec} \left(\frac{x}{2}\right)$
$\tan \left(\operatorname {arcsec} \left(\frac{x}{2}\right)\right) - \left(\operatorname {arcsec} \left(\frac{x}{2}\right)\right) + c$
This is wrong but I am not sure why.