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#