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Math Help - Unusual Integral

  1. #1
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    Unusual Integral

    Evaluate

    \int \frac{1}{e^x + 1} dx.
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  2. #2
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    -ln(e^x+1)+x
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  3. #3
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     \frac{1}{e^x + 1} \times \frac{e^{-x}}{e^{-x}}= \frac{e^{-x}}{1 + e^{-x}}

    Can you go from here?
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  4. #4
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    Quote Originally Posted by icemanfan View Post
    Evaluate

    \int \frac{1}{e^x + 1} dx.
    = \int \frac{e^x + 1 - e^x}{e^x + 1} dx

    = \int \frac{e^x + 1}{e^x + 1}dx - \int \frac{e^x}{e^x + 1} dx

    = x - \ln(e^x + 1) + c
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  5. #5
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    \int \frac{1}{1 + e^x}dx = \int \frac{e^{-x}}{1 + e^{-x}} dx = -\ln(1 + e^{-x}) + C

    Correct?
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  6. #6
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    Quote Originally Posted by icemanfan View Post
    \int \frac{1}{1 + e^x}dx = \int \frac{e^{-x}}{1 + e^{-x}} dx = -\ln(1 + e^{-x}) + C

    Correct?
    You lost an x and your e should be to the positive power.
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  7. #7
    Super Member Deadstar's Avatar
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    Quote Originally Posted by dwsmith View Post
    You lost an x and your e should be to the positive power.
    Nah it's right I think.

    Someone show me how the two answers we have given are equiv though..?
    Last edited by Deadstar; April 12th 2010 at 05:29 PM.
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  8. #8
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    Quote Originally Posted by Deadstar View Post
    Nah it's right I think.

    Someone show me how the two answers we have given are equiv though..?
    Yeah they are equivalent. I checked them in Maple; however, I have no idea how to show that they are.
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  9. #9
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    Quote Originally Posted by Deadstar View Post
    Nah it's right I think.

    Someone show me how the two answers we have given are equiv though..?
    -ln(1-e^-x) = -ln(1 - 1/e^x) = -ln[(e^x - 1)/e^x]
    = -[ln(e^x - 1) - lne^x]
    = x- ln(e^x - 1)
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