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

Nov 13, 2015

First let's factor the denominator:

${\left({x}^{2} - 1\right)}^{2}$

we see the difference of squares, so we can break that down like so...
${\left(\left(x - 1\right) \left(x + 1\right)\right)}^{2}$

Now we can "distribute" the power to the two terms since they are being multiplied
${\left(x - 1\right)}^{2} {\left(x + 1\right)}^{2}$

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

$\frac{2 {x}^{2} - x}{{\left(x - 1\right)}^{2} {\left(x + 1\right)}^{2}} = \frac{A}{x - 1} + \frac{B}{x - 1} ^ 2 + \frac{C}{x + 1} + \frac{D}{x + 1} ^ 2$

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

$2 {x}^{2} - x = A \left(x - 1\right) {\left(x + 1\right)}^{2} + B {\left(x + 1\right)}^{2} + C \left(x + 1\right) {\left(x - 1\right)}^{2} + D {\left(x - 1\right)}^{2}$

To find the constants, let's just plug something in for $x$.
let $x = 1$
$2 - 1 = \cancel{A \left(0\right) {\left(2\right)}^{2}} + B {\left(2\right)}^{2} + \cancel{C \left(2\right) {\left(0\right)}^{2}} + \cancel{D {\left(0\right)}^{2}}$
$1 = B \left(4\right)$
$B = \frac{1}{4}$

Now let $x = - 1$

$2 + 1 = \cancel{A \left(- 2\right) {\left(0\right)}^{2}} + \cancel{\left(\frac{1}{4}\right) {\left(0\right)}^{2}} + \cancel{C \left(0\right) {\left(- 2\right)}^{2}} + D {\left(- 2\right)}^{2}$
$3 = D \left(4\right)$
$D = \frac{3}{4}$

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

$2 {x}^{2} - x = A \left({x}^{3} + {x}^{2} - x - 1\right) + \frac{1}{4} \left({x}^{2} + 2 x + 1\right) + C \left({x}^{3} - {x}^{2} - x + 1\right) + \frac{3}{4} \left({x}^{2} - 2 x + 1\right)$

$2 {x}^{2} - x = A {x}^{3} + A {x}^{2} - A x - A + \frac{1}{4} {x}^{2} + \frac{1}{2} x + \frac{1}{4} + C {x}^{3} - C {x}^{2} - C x + C + \frac{3}{4} {x}^{2} - \frac{3}{2} x + \frac{3}{4}$

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...

$0 {x}^{3} + 2 {x}^{2} - x + 0$
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

$0 {x}^{3} = A {x}^{3} + \cancel{A {x}^{2} - A x - A + \frac{1}{4} {x}^{2} + \frac{1}{2} x + \frac{1}{4}} + C {x}^{3} - \cancel{C {x}^{2} - C x + C + \frac{3}{4} {x}^{2} - \frac{3}{2} x + \frac{3}{4}}$

$0 {x}^{3} = A {x}^{3} + C {x}^{3}$
so $A + C = 0$

now let's do the constants...

$0 = \cancel{A {x}^{3} + A {x}^{2} - A x} - A + \cancel{\frac{1}{4} {x}^{2} + \frac{1}{2} x} + \frac{1}{4} + \cancel{C {x}^{3} - C {x}^{2} - C x} + C + \cancel{\frac{3}{4} {x}^{2} - \frac{3}{2} x} + \frac{3}{4}$
$0 = - A + \frac{1}{4} + C + \frac{3}{4}$
$0 = 1 + C - A$
$- 1 = C - A$

since we know the following :
$0 = C + A$
$- 1 = C - A$
you should be able to figure out $C = - \frac{1}{2}$ and $A = \frac{1}{2}$ (just substitute)

now we can finally rewrite the integral like so:

$\int \left(\frac{1}{2 \left(x - 1\right)} + \frac{1}{4 {\left(x - 1\right)}^{2}} - \frac{1}{2 \left(x + 1\right)} + \frac{3}{4 {\left(x + 1\right)}^{2}}\right) \mathrm{dx}$

These two parts should be easy...
$\frac{1}{2} \int \frac{1}{x - 1} \mathrm{dx} - \frac{1}{2} \int \frac{1}{x + 1} \mathrm{dx}$
$\frac{1}{2} \ln | x - 1 | - \frac{1}{2} \ln | x + 1 |$

Now the other two...
$\frac{1}{4} \int \frac{1}{{\left(x - 1\right)}^{2}} \mathrm{dx} + \frac{3}{4} \int \frac{1}{{\left(x + 1\right)}^{2}} \mathrm{dx}$

$\frac{1}{4} \cdot \left(- \frac{1}{x - 1}\right) + \frac{3}{4} \left(- \frac{1}{x + 1}\right)$

all together we have:

$\frac{1}{2} \ln | x - 1 | - \frac{1}{2} \ln | x + 1 | - \frac{1}{4 \left(x - 1\right)} - \frac{3}{4 \left(x + 1\right)} + C$