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Why do you Tack on The Negative For this Integral

  1. Feb 7, 2012 #1
    If you integrate

    1/(1-y)dy

    why do you end up with a negative in front of your answer

    -ln|1-y|+c
     
  2. jcsd
  3. Feb 7, 2012 #2

    Char. Limit

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    Gold Member

    Because using u-substitution, we get this:

    u = 1 - y, du = - dy

    And so...

    [tex]\int \frac{1}{1-y} dy = - \int \frac{1}{u} du[/tex]

    Integrating the right-hand side, we get - ln|u| + C, or - ln|1-y| + C.
     
  4. Feb 7, 2012 #3
    gotcha
     
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