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Math Help - partial fraction decomposition help

  1. #1
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    partial fraction decomposition help

    \frac{1}{(x^2+1)(x+3)^2}

    Heres my attempt:

    \frac{1}{(x^2+1)(x+3)^2}=\frac{Ax+B}{x^2+1}+\frac{  C}{x+3}+\frac{D}{(x+3)^2}

    1=(Ax+B)(x+3)^2+C(x^2+1)(x+3)+D(x^2+1)

    1=Ax^3+6Ax^2+9Ax+Bx^2+6Bx+9B+Cx^3+3Cx^2+Cx+3C+Dx^2  +D

    9B+3C+D=1
    9A+6B+C=0
    6A+B+3C+D=0
    A+C=0

    This is what I get for A, B C, and D
    A=\frac{1}{50}

    B=\frac{7}{50}

    C=- \frac{1}{50}

    D=- \frac{1}{5}

    Where did I go wrong?
    Last edited by yoman360; August 28th 2010 at 11:35 AM.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    It seems you missed something while solving simultaneously...

    Anyway, if you take from here:

        1=(Ax+B)(x+3)^2+C(x^2+1)(x+3)+D(x^2+1)

    Putting x = -3 cancels A, B and C all at once, giving:

    1 = D((-3)^2 +1)

    And

    D = \frac{1}{10}
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  3. #3
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    your right i made one tiny mistake which messed up the whole thing. >.<
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  4. #4
    MHF Contributor Unknown008's Avatar
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    I agree. One slight mistake can ruin the whole thing since there are so many 'variables' to find...

    Anyway, I then used wolframalpha and got the partial fractions:

    http://www.wolframalpha.com/input/?i=\frac{1}{(x^2%2B1)(x%2B3)^2}
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