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Partial Fractions problem

  1. Feb 24, 2014 #1
    1. The problem statement, all variables and given/known data

    Use integration by parts to evaluate the integral

    ∫(7-6x) / (x2-4x+13)


    3. The attempt at a solution
    This is a question from my notes so I already have the solution but I'm not sure what's going on at this one specific step.

    ∫(7-6x) / (x2-4x+13)

    = -∫(6x-7) / (x2-4x+13)

    = -∫( 3(2x-4) + 5) / (x2-4x+13)

    So I know 2x-4 is the derivative of the denominator but I don't know why the numerator is changing to 3(2x-4) + 5.
     
    Last edited by a moderator: Feb 24, 2014
  2. jcsd
  3. Feb 24, 2014 #2

    Mark44

    Staff: Mentor

    Please don't use the SIZE tags. What you wrote is legible enough without making it larger.

    Ignoring the integration part for the moment, here's the algebra. What they're doing is writing the original fraction as
    $$\frac{-3(2x - 4) - 5}{x^2 - 4x + 13} = \frac{-3(2x - 4)}{x^2 - 4x + 13} + \frac{-5}{x^2 - 4x + 13}$$

    Now as an integration problem, the first fraction can be integrated using an ordinary substitution and the second can be done using a formula you've probably already seen.

    Your title is misleading - I don't see this as a partial fractions problem. I also don't see this as in integration by parts problem, either.
     
  4. Feb 24, 2014 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The numerator changes to -3[(2x-4) + 5], which is equal to 7-6x.

    ehild
     
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