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Math Help - indefinite integral #2

  1. #1
    Super Member Random Variable's Avatar
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    indefinite integral #2

    Challenge Problem:

     \int \frac{x+x^{2}}{(1+x+e^{x})^{2}} \ dx

    Moderator approved CB
    Last edited by CaptainBlack; December 26th 2010 at 10:06 PM.
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  2. #2
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     \int \frac{x^2+x}{(1+x+e^x)^2} ~dx

    Sub.  u = 1+e^{-x}(1+x)

     du = e^{-x}(1 - 1 - x ) dx = - xe^{-x} dx

    The integral

     = - \int \frac{e^{-x}(1+x)}{(1+e^{-x}(1+x) )^2} ( -xe^{-x}~dx)

     = - \int \frac{ u-1}{u^2}~du

     = - \ln|u| - \frac{1}{u} + C

     = - \ln| 1+e^{-x}(1+x)| - \frac{e^x }{1+x+e^x} + C

     = x - \ln|1+x+e^x| - \frac{e^x}{1+x+e^x} + C

    EDIT: Surely , they differ by a constant  \frac{x+1}{1+x+e^x} - ( - \frac{e^x}{1+x+e^x} ) = \frac{1+x+e^x}{1+x+e^x} = 1
    Last edited by simplependulum; December 26th 2010 at 09:51 PM.
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  3. #3
    Super Member Random Variable's Avatar
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    Unless  \frac{x+1}{1+x+e^{x}} and  -\frac{e^{x}}{1+x+e^{x}} differ by a constant, something is wrong with your solution. Perhaps the way you rewrote the integrand.
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  4. #4
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    Quote Originally Posted by Random Variable View Post
    Unless  \frac{x+1}{1+x+e^{x}} and  -\frac{e^{x}}{1+x+e^{x}} differ by a constant, something is wrong with your solution. Perhaps the way you rewrote the integrand.
    \displaystyle \left(\frac{x+1}{1+x+e^{x}}\right)' = -\frac{xe^x}{(x+e^x+1)^2} = \left(-\frac{e^{x}}{1+x+e^{x}}\right)', so they do differ by a constant.
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