# 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.

• If I substitute $\frac{x^2-1}{x^2-4}$ with $t^2$ then calculating $dt$ results in a hard-to-simplify derivative. I don't see how I could move on afterwards. Could you please show some extra steps? – Vittorio Romeo Jan 31 '14 at 17:10
• Under that substitution, what is the relation between $dx$ and $dt$ so one could finish the substitution? – coffeemath Jan 31 '14 at 18:35
• Може ли да ви отговоря на български? – kmitov Jan 31 '14 at 20:34
• At the last step, you go from an expression for $x^2$ right to $dx/dt$. However the derivative of $x^2$ gives $2x(dx/dt)$ which is not what you have. – coffeemath Feb 1 '14 at 5:27