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Integrating 1/(1+e^x)

  1. Jul 8, 2003 #1
    Integrating 1/(1+e^x):

    I am currently substituting 1+e^x = u,

    S (u*e^x)^-1 du since du/dx = e^x

    S (u*(u-1))^-1 du (S represents the integral symbol)

    But i looked up the answer and got Ln (e^x(1+e^x)) +c

    i need to understand how this is done, thanks in advance everyone.
     
  2. jcsd
  3. Jul 8, 2003 #2
  4. Jul 8, 2003 #3

    HallsofIvy

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    After you have integral (u(u-1))^(-1) du= integral 1/(u(u-1)) du

    you can use "partial fractions", rewriting

    1/(u(u-1))= A/u+ B/(u-1) for proper A and B. Then integrate

    integral (A/u)du+ integral B/(u-1) du.
     
  5. Jul 23, 2008 #4
    Guys this problem is a lot easier than it looks. From a quote from a PHD Student at Texas Tech Univ. Krishna Kaphle "Adding 0 or Multiplying By 1 is Free In Math".

    What does this mean for this problem well lets rewrite our expression
    1. [tex]\int{\frac{1+e^x-e^x}{1+e^x}dx}[/tex] [lets add 1 to our expression]

    2. [tex]\int{\frac{[1+e^x]-e^x}{1+e^x}dx}[/tex] [now focus in on the numerator see whats in brackets matches the denominator]

    3. [tex]\int{\frac{1+e^x}{1+e^x}dx}-\int{\frac{e^x}{1+e^x}dx}[/tex] [now lets split the fraction]

    4. [tex]\int{1dx}-\int{\frac{e^x}{1+e^x}dx}[/tex] [now just use substiution method for right hand integral. [tex]u=1+e^x, du = e^xdx[/tex]]

    5. [tex]\int{1dx}-\int{\frac{du}{u}}[/tex]

    6. [tex]x-ln|{u}|+C[/tex]

    7. [tex]x-ln|{1+e^x}|+C[/tex] [substitue u back]

    If you don't believe this or too lazy to do its derivative use this tool:
    http://library.wolfram.com/webMathematica/Education/WalkD.jsp
    and enter this expression: x-ln[1+e^x]

    kiko
     
  6. Dec 1, 2008 #5
    If the poster above can bump this thread, so can I. Besides, I think my solution is the nicest one between this thread and the thread referred to in post #2. :)

    I make the substitution [tex]u = e^{-x} + 1[/tex]:
    [tex]
    \int \frac{1}{1 + e^x} \,dx
    = \int \frac{e^{-x}}{e^{-x} + 1} \,dx
    = \int -\frac{1}{u} \,du
    = -\ln \lvert u \rvert + C
    = -\ln(e^{-x} + 1) + C.
    [/tex]

    Note that
    [tex]-\ln(e^{-x} + 1) = x - \ln(1 + e^x) = \ln \left( \frac{e^x}{1 + e^x} \right).[/tex]


    edit: Isn't this thread supposed to be locked, being archived?
     
    Last edited: Dec 1, 2008
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