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Thread: Partial Fractions Help

  1. #1
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    Partial Fractions Help

    $\displaystyle \int\frac{x-6}{x^2(x-2)}$

    $\displaystyle \frac{x-6}{x^2(x-2)}=\frac{A}{x-2}+\frac{B}{x}+\frac{C}{x^2}$

    $\displaystyle x-6=Ax^2+Bx^2-2B+Cx-2C$

    $\displaystyle x-6=x^2(A+B)+Cx-2(B+C)$

    $\displaystyle 3=B+C$

    $\displaystyle 0=A+B$

    $\displaystyle 1=C$

    $\displaystyle A=-2$

    $\displaystyle B=2$

    $\displaystyle C=1$

    I know this is wrong; where have I screwed up. Using the cover up method, I know that A should equal -1. Help. TIA.
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  2. #2
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    Re: Partial Fractions Help

    Hey wondering.

    I think your line should be x - 6 = Ax^2 + Bx(x-2) + C(x-2) which gives:

    x - 6 = Ax^2 + Bx^2 - 2Bx + Cx - 2C which has a term 2Bx instead of 2B if you are using that particular decomposition listed above.
    Thanks from wondering
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  3. #3
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    Re: Partial Fractions Help

    Thank you. It must be time to quit doing math for the night.
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