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 $\sqrt{x^2-4}= \tan u$?
 A: 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:
$$\int\frac{\sqrt{x^2-4}}{x}dx=\sqrt{x^2-4}-2\arctan\frac{\sqrt{x^2-4}}{2}+C$$
A: The substitution $x=2\sec t$ works well.
You will end up wanting to integrate a constant times $\tan^2 t$. Rewrite $\tan^2 t$ as $\sec^2 t-1$.  
Remark: There are other approaches, for example the hyperbolic function substitution of M. Strochyk. 
Or else we can rewrite our integrand as $\frac{x\sqrt{x^2-4}}{x^2}$. Then let $x^2-4=u^2$. So $2x\,dx=2u\,du$. We end up wanting $\int \frac{u^2\, du}{u^2+4}$. Rewrite $\frac{u^2}{u^2+4}$ as $1-\frac{4}{u^2+4}$. 
Another substitution that works fairly nicely here is $x=\frac{1}{t}$. 
In a North American first calculus course, $x=2\sec t $ would be the standard expected approach.  
A: Hint:
Try substitution $x=2\cosh{u}.$ Then $$x^2-4=4(\cosh^2{u}-1)=4\sinh^2{u}, \;\; dx=2\sinh{u} \ du,$$
hence
$$\int\frac{\sqrt{x^2-4}}{x}dx=\int \frac{2 \sinh{u} \cdot 2\sinh{u} \ du}{\cosh{u}}=4\int \frac{ \cosh^2{u}-1 }{\cosh{u}}\  du= \\ 
=4 \int{\cosh{u} \ du} - 4\int{\frac{du}{\cosh{u}}}=4\sinh{u}-4\int{\frac{2 \ du}{e^u+e^{-u}}}= \\ 
= 4\sinh{u}-8\int{\frac{e^u \ du}{e^{2u}+1}}=4\sinh{u}-8\arctan{e^u}.$$
A: As the expression goes, there is more than one way to skin a cat. (Sorry, PETA.)
I'd say the substitution $x=2\sec t$ is the more obvious choice.  It eliminates the square root (after making use of the identity $\sec^2t-1=\tan^2 t$), should leave something somewhat manageable, and resubstitution shouldn't be that bad.
The substitution $u=\sqrt{x^2-4}$ is less obvious.  If you rewrite the integral as
$$\int\frac{x(x^2-4)dx}{x^2\sqrt{x^2-4}}=\int\big(\frac{x^2-4}{x^2}\big)\big(\frac{xdx}{\sqrt{x^2-4}}\big)$$
you can now see that as 
$$\int\frac{u^2du}{u^2+4}=\int du-4\int\frac{du}{u^2+4}$$
which should lead to a solution.
