# Help! Partial Fractions problem.

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• Sep 19th 2010, 06:21 PM
Bracketology
Help! Partial Fractions problem.
[IMG]file:///C:/Users/MICHAE%7E1/AppData/Local/Temp/moz-screenshot.png[/IMG]Attachment 18981

Not really sure how to do this one..Or even if the problem is written correctly.
• Sep 19th 2010, 06:42 PM
Archie Meade
Quote:

Originally Posted by Bracketology
[IMG]file:///C:/Users/MICHAE%7E1/AppData/Local/Temp/moz-screenshot.png[/IMG]Attachment 18981

Not really sure how to do this one..Or even if the problem is written correctly.

$\displaystyle\frac{15}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}=\frac{x+1}{x+1}\ \frac{A}{x-1}+\frac{x-1}{x-1}\ \frac{B}{x+1}$

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

Therefore $15=15+(0)x=(A-B)+(A+B)x$

which gives us 2 simultaneous equations for A and B

$A+B=0\Rightarrow\ A=-B$

$A-B=15\Rightarrow\ A+A=15$

Get the idea?
• Sep 19th 2010, 06:43 PM
skeeter
factor the constant 15 out from the integral ...

$\displaystyle \frac{1}{x^2-1} = \frac{A}{x-1} + \frac{B}{x+1}$

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

$1 = A(x+1) + B(x-1)$

let $x = 1$ ... $\displaystyle A = \frac{1}{2}$

let $x = -1$ ... $\displaystyle B = -\frac{1}{2}$

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

$\displaystyle \frac{15}{2} \int \frac{1}{x-1} - \frac{1}{x+1} \, dx
$

now integrate.
• Sep 19th 2010, 06:48 PM
Bracketology
Thanks, I appreciate the help.