Integral of square root of a fraction of two functions $$\int\sqrt{\frac{x^2-1}{x^2-4}}~dx$$
I'm having a lot of trouble solving this integral. I can't seem to find any way to simplify it.
I tried to split the integral in two, but I couldn't find a way. I tried to find something that would let me have $\dfrac{f'(x)}{f(x)}$, but I had no luck.
Any hint?

The substitution $\frac{x^2-1}{x^2-4}=t^2$ will transform the integrand into a rational fraction.
$x^2-1=t^2(x^2-4)$
$x^2-1=x^2t^2-4t^2$
$x^2(1-t^2)=1-4t^2$
$x^2=\frac{4t^2-1}{t^2-1}$
$\frac {dx}{dt}=\frac{8t(t^2-1)-2t(4t^2-1)}{(t^2-1)^2}dt=\frac{10t}{(t^2-1)^2}dt$
Now the integral is $\int\frac{10 t^2}{(t^2-1)^2}dt$ and you can use partial fractions.
