Help with $\large \intop\frac{\sqrt{x^2-4}}{x} dx$ I have to $\large \intop\frac{\sqrt{x^2-4}}{x} dx$. Can you tell me what to substitute? 
Should I substitute $x$ or $\sqrt{x^2-4}$? Would it be better If I'd substitute $x=\tan u$
or should I substitute $\sqrt{x^2-4}= \tan u$?

You may try several substitutions, perhaps more elementary and maybe easier:
$$(1)\;\;\text{First substitution :}\;\;u=x^2-4\Longrightarrow du=2xdx\Longrightarrow dx=\frac{du}{2\sqrt{u+4}}\;\;,\;\;\text{so :}$$
$$\int\frac{\sqrt{x^2-4}}{x}dx=\int\frac{\sqrt u}{2(u+4)}du$$
$$(2)\;\;\text{Second substitution :}\;\;\;\;\;\;\;\;y^2=u\Longrightarrow 2ydy=du\;\;,\;\;\text{so :}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$$
$$\int\frac{\sqrt u}{2(u+4)}du=\int\frac{y^2}{y^2+4}dy=\int\left(1-\frac{4}{y^2+4}\right)dy=$$
$$=y+2\int\frac{\frac{1}{2}dy}{1+\left(\frac{y}{2}\right)^2}=y+\arctan\frac{y}{2}+C$$
And now go back to your original variable to get:
$$\int\frac{\sqrt{x^2-4}}{x}dx=\sqrt{x^2-4}-2\arctan\frac{\sqrt{x^2-4}}{2}+C$$
