Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$\int\sqrt{\frac{x^2-1}{x^2-4}}~dx$$

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?

share|improve this question
2  
en.wikipedia.org/wiki/Elliptic_integral –  user114628 Jan 31 at 17:02

1 Answer 1

up vote 0 down vote accepted

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.

share|improve this answer
2  
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 at 17:10
    
Under that substitution, what is the relation between $dx$ and $dt$ so one could finish the substitution? –  coffeemath Jan 31 at 18:35
1  
Може ли да ви отговоря на български? –  kmitov Jan 31 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 at 5:27

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.