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Integral calculation help

  1. Feb 24, 2006 #1
    I'm working on thiis:
    [tex].25 = \int_{-\infty}^m\frac{e^{x}}{(1+e^{-x})^2}dx[/tex]
    I let [tex]u=1+e^{-x} , du=-e{-x}dx[/tex]
    which gives:
    then I solve it out and I get
    I have tried to solve this and I keep getting a negative ln. I moved the 1 to the RHS and then took ln of both sides. I can't get it to work.
  2. jcsd
  3. Feb 24, 2006 #2


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    Homework Helper

    But you don't have an e-x in the numerator. Try multiplying the top and bottom by e2x, then using the substitution u=1+ex, which will leave you with something like (u-1) to some power divided by u2.

    By the way, there is no solution to the equation you mention at the end since 1/(1+ex)<1 for any real x.
  4. Feb 24, 2006 #3
    Statusx, you are absolutely correct. I have forgotten my - sign. It should read

    [tex].25 = \int_{-\infty}^m\frac{e^{-x}}{(1+e^{-x})^2}dx[/tex]
    then let

    [tex]u=1+e^{-x} , du=-e^{-x}dx[/tex]
    which gives
    so then you get
    from there I took to ln of both sides and tried to solve for m, but nothing is working out.
  5. Feb 24, 2006 #4
    ln makes no sense
  6. Feb 24, 2006 #5


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    Gold Member
    Dearly Missed

    It makes no sense because you've been sloppy! :grumpy:
    We have:
  7. Feb 24, 2006 #6
    der duh der duh
    I am a sloppy, sloppy student.:redface:
    I'm all good now.
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