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.





$\frac {dx}{dt}=\frac{8t(t^2-1)-2t(4t^2-1)}{(t^2-1)^2}dt=\frac{10t}{(t^2-1)^2}dt$

Now the integral is $\int\frac{10 t^2}{(t^2-1)^2}dt$ and you can use partial fractions.

  • 2
    $\begingroup$ 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? $\endgroup$ – Vittorio Romeo Jan 31 '14 at 17:10
  • $\begingroup$ Under that substitution, what is the relation between $dx$ and $dt$ so one could finish the substitution? $\endgroup$ – coffeemath Jan 31 '14 at 18:35
  • 1
    $\begingroup$ Може ли да ви отговоря на български? $\endgroup$ – kmitov Jan 31 '14 at 20:34
  • $\begingroup$ 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. $\endgroup$ – coffeemath Feb 1 '14 at 5:27

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