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Math Help - Help! Partial Fractions problem.

  1. #1
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    Help! Partial Fractions problem.

    [IMG]file:///C:/Users/MICHAE%7E1/AppData/Local/Temp/moz-screenshot.png[/IMG]Help! Partial Fractions problem.-problem.png

    Not really sure how to do this one..Or even if the problem is written correctly.
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  2. #2
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    Quote Originally Posted by Bracketology View Post
    [IMG]file:///C:/Users/MICHAE%7E1/AppData/Local/Temp/moz-screenshot.png[/IMG]Click image for larger version. 

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    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?
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  3. #3
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    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<br />

    now integrate.
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  4. #4
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    Thanks, I appreciate the help.
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