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Derivatives of partial fractions

  1. Oct 14, 2010 #1
    I'm having issue with one problem. We're asked to break down the problem into partial fractions to solve for the integral.

    Well, I'm stuck on one. I'm being asked for the values of A, B, and C for the following problem.

    ∫((9x^2+13x-83)/((x-3)(x^2 + 16)))dx

    I can get it worked down pretty far, but I'm continually given back that I've done it incorrectly.

    Broken down:

    (9x^2+13x-83)/((x-3)(x^2 + 16)) = A/(x-3) + (bx + c)/(x^2 + 16)

    From there, I get (multiplying both sides by the common denominator, and now only working with the right side as it's the only one that needs to be worked):

    A(x^2+12) + Bx(x-3) + C(x-3)
    = ax^2 + 12A + Bx^2 - 3Bx + Cx - 3C
    = (A+B)x^2 + (C-3B)x + (12A - 3C)


    -9x^2 + 13x -83 = (A+B)x^2 + (C-3B)x + (12A - 3C)

    Breaking it down further, I get:

    A + B = -9
    3B - C = 13
    12A - 3C = -83

    I'm thinking something may be wrong with my values in the second part. Any help would be appreciated. Whatever answers I put in, I'm marked as incorrect. I went so far as to plug the last little part I posted into Wolfram|Alpha to confirm, and it spits the same answers I've been getting. I get the feeling that I did something incorrectly earlier on.
  2. jcsd
  3. Oct 15, 2010 #2
    I think the mistake is here, it should be A(x^2+16)
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