How do you find `int (2x^2 -x)/((x^2 -1)^2)dx` using partial fractions?

First let's factor the denominator:

`(x^2-1)^2`

we see the difference of squares, so we can break that down like so...
`((x-1)(x+1))^2`

Now we can "distribute" the power to the two terms since they are being multiplied
`(x-1)^2(x+1)^2`

Now we see two repeated linear terms, so our partial fraction decomposition will look like this:

`(2x^2-x)/((x-1)^2(x+1)^2)=A/(x-1)+B/(x-1)^2+C/(x+1)+D/(x+1)^2`

After getting common denominators you should get the following (Sorry, I kinda skipped a step):

`2x^2-x=A(x-1)(x+1)^2+B(x+1)^2+C(x+1)(x-1)^2+D(x-1)^2`

To find the constants, let's just plug something in for `x`.
let `x=1`
`2-1=cancel(A(0)(2)^2)+B(2)^2+cancel(C(2)(0)^2)+cancel(D(0)^2)`
`1=B(4)`
`B=1/4`

Now let `x=-1`

`2+1=cancel(A(-2)(0)^2)+cancel((1/4)(0)^2)+cancel(C(0)(-2)^2)+D(-2)^2`
`3=D(4)`
`D=3/4`

We might need some other trickery to find `A` and `C`...
let's try multiplying the `x`-terms out here (this will get messy)
`2x^2-x=A(x-1)(x+1)^2+1/4(x+1)^2+C(x+1)(x-1)^2+3/4(x-1)^2`

`2x^2-x=A(x^3+x^2-x-1)+1/4(x^2+2x+1)+C(x^3-x^2-x+1)+3/4(x^2-2x+1)`

`2x^2-x=Ax^3+Ax^2-Ax-A+1/4x^2+1/2x+1/4+Cx^3-Cx^2-Cx+C+3/4x^2-3/2x+3/4`

This is a MESS to sort through, but a quick thing we can do is look at the coefficients in front of the powers on the left side...

`0x^3+2x^2-x+0`
These have to match the coefficients in front of the power on the right side, so we'll collect up only the coefficients in front of one at a time...
let's do `x^3` first

`0x^3=Ax^3+cancel(Ax^2-Ax-A+1/4x^2+1/2x+1/4)+Cx^3-cancel(Cx^2-Cx+C+3/4x^2-3/2x+3/4)`

`0x^3 = Ax^3 + Cx^3`
so `A+C=0`

now let's do the constants...

`0 = cancel(Ax^3+Ax^2-Ax)-A+cancel(1/4x^2+1/2x)+1/4+cancel(Cx^3-Cx^2-Cx)+C+cancel(3/4x^2-3/2x)+3/4`
`0 = -A + 1/4 +C +3/4`
`0 = 1+C-A`
`-1 = C-A`

since we know the following :
`0=C+A`
`-1 =C-A`
you should be able to figure out `C=-1/2` and `A=1/2` (just substitute)

now we can finally rewrite the integral like so:

`int (1/(2 (x-1))+1/(4 (x-1)^2)-1/(2 (x+1))+3/(4 (x+1)^2)) dx`

These two parts should be easy...
`1/2 int 1/(x-1) dx-1/2 int 1/(x+1) dx`
`1/2 ln|x-1| - 1/2 ln|x+1|`

Now the other two...
`1/4int 1/((x-1)^2)dx+3/4int1/( (x+1)^2) dx`

`1/4*(-1/(x-1))+3/4(-1/(x+1))`

all together we have:

`1/2 ln|x-1| - 1/2 ln|x+1| -1/(4(x-1))-3/(4(x+1)) +C`