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Indefinite Integration of a Rational Expression

  1. Aug 12, 2012 #1
    1. The problem statement, all variables and given/known data
    <Indefinite integral sign here>[r^2 -2r] / [r^3 - 3r^2 + 1]dr
    or the second example in the "Substitution" section here:
    http://people.clarkson.edu/~sfulton/ma132/parfrac.pdf

    2. Relevant equations
    nada.


    3. The attempt at a solution
    Nothing to really attempt, I just don't get what they do with the numerator (i.e., how it turns into a 1).
     
  2. jcsd
  3. Aug 12, 2012 #2

    tiny-tim

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    hi communitycoll! :smile:
    (have an integral: ∫ and try using the X2 button just above the Reply box :wink:)
    ah, nooo, the numerator isn't r2 - 2r, it's (r2 - 2r)dr !!

    (and of course that's 1 times dw :wink:)

    the trick in substitution is that you always have to substitute the "d" part also! :smile:
     
  4. Aug 12, 2012 #3

    eumyang

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    If you read that attachment carefully, you would have seen this:
    So they let
    w = r3 - 3r2 + 1,
    so
    dw = 3r2 - 6r dr,
    which is the same as
    dw = 3(r2 - 2r) dr.

    Do you see it now?
     
  5. Aug 14, 2012 #4
    Okay then. Yeah, I understand now. Thanks.
     
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