Integral of $\int \frac {\sqrt {x^2 - 4}}{x} dx$ 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.

\begin{aligned}
& \int \frac{\sqrt{x^{2}-4}}{x} d x \\
=& \int \frac{x^{2}-4}{x \sqrt{x^{2}-4}} d x \\
=& \int \frac{x^{2}-4}{x^{2}} d\left(\sqrt{x^{2}-4}\right) \\
=& \sqrt{x^{2}-4}-4\int{\frac{1}{x^{2}} d\left(\sqrt{x^{2}-4}\right)} \\
=& \sqrt{x^{2}-4}-4\int \frac{d\left(\sqrt{x^{2}}-4\right)}{\left(\sqrt{x^{2}-4}\right)^{2}+4}\\
=& \sqrt{x^{2}-4}-2\tan ^{-1}\left(\frac{\sqrt{x^{2}-4}}{2}\right)+C
\end{aligned}
