1. ## Integration-Partial Fractions

$
\int {\frac{{dx}}
{{(x^2 - 1)^2 }}} = \int {\frac{{dx}}
{{(x - 1)^2 (x + 1)^2 }}}
$

It seems no matter what I try after this I cannot get the answer in the book.
Which is:

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

Any help would be greatly appreciated.

2. Originally Posted by kid funky fried
$
\int {\frac{{dx}}
{{(x^2 - 1)^2 }}} = \int {\frac{{dx}}
{{(x - 1)^2 (x + 1)^2 }}}
$

It seems no matter what I try after this I cannot get the answer in the book.
Which is:

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

Any help would be greatly appreciated.
$\frac{1}{(x^2-1)^2}=\frac{1}{(x-1)^2(x+1)^2}$

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

Clearing the fractions gives

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

First lets get our two freebee values

let x =1

$1=A(1-1)(x+1)^2+B(1+1)^2+C(1-1)^2(1+1)+D(1-1)^2$
$1=b(2)^2 \iff \frac{1}{4}=B$

x=-1 gives

$1=D(-1-1)^2 \iff \frac{1}{4}=D$

Now we know that this is true for ALL VALUES OF X so we can pick some others. I will pick x=0 and x=2 to get a system of equations.

$1=A(0-1)(0+1)^2+\frac{1}{4}(0+1)^2+C(0-1)^2(0+1)+\frac{1}{4}(0-1)^2$

$1=-A+\frac{1}{4}+C+\frac{1}{4} \iff \frac{1}{2}=-A+C$

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

$1=9A+\frac{9}{4}+3C+\frac{1}{4} \iff -\frac{3}{2}=9A+3C \iff -\frac{1}{2}=3A+C$

Now we need to solve
$3A+C=-\frac{1}{2}$
$-A+C=\frac{1}{2}$

subtracing the 2nd from the first gives

$4A=-1 \iff A=-\frac{1}{4}$

then $-\left( -\frac{1}{4}\right)+C=\frac{1}{2} \iff C=\frac{1}{4}$

There we go Yeah!!!

I hope this helps.

3. Yes it helps alot. I had gotten as far as finding B and D. A and C had me perplexed.
Thanks, I appreciate it.