Question

How do you find the integral of `dx / (x^2 - 4)^2`?

Answer 1

Quick answer :

`int1/(x^2-4)^2dx = 1/16int1/(-1/4x^2+1)dx`

substitute `u = 1/2x`

`du = 1/2`

`u^2 = 1/4x^2`

So `1/8int 1/(-u^2+1)^2du`

Here you can do partial fraction but it's long...

let's `u = tanh(t)`

`du = 1/cosh^2(t)dt`

`-u^2 = -tanh^2(t)`

don't forget `1-tanh^2(t) = 1/cosh^2(t)`

so we have

`1/8int1/(1/cosh^2(t))^2*1/cosh^2(t)dt = 1/8intcosh^2(t)`

dont forget `cosh^2(t) = 1/2(1+cosh(2t))`

`1/16int1+cosh(2t)dt`

`1/16[t+1/2sinh(2t)]+C`

and then substitute back