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Integration Y= (1/(e^x + 1))

  1. Dec 23, 2005 #1
    I seem to be having trouble with this antiderivative can i please see a solution? Y= (1/(e^x + 1))

    I tried substituting e^x + 1 as u, but that left me with the quest to find an antiderivative of 1/(u^2 - u). i feel like im missing something silly here. thanks
     
  2. jcsd
  3. Dec 23, 2005 #2

    benorin

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    You have it, keep going use partial fractions: [itex] \frac{1}{u(u-1)}= -\frac{1}{u}+\frac{1}{u-1}[/itex]
     
  4. Dec 23, 2005 #3
    Try multiplying top and bottom by e^(-x)
     
  5. Jan 4, 2006 #4

    benorin

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    Here it is (the integral)

    For

    [tex]\int\frac{dx}{e^x +1}[/tex]

    put [itex]u=e^x +1\Rightarrow x=\log (u-1) \mbox{ so that }dx=\frac{du}{u-1}[/itex]

    to get

    [tex]\int\frac{dx}{e^x +1} = \int\frac{du}{u(u-1)} = \int\left( -\frac{1}{u}+\frac{1}{u-1}\right) du = -\log |u| + \log |u-1| + C [/tex]
    [tex]= -\log \left| e^x +1\right| + \log \left| e^x\right| + C = -\log \left| e^x +1\right| + x + C[/tex]
     
  6. Jan 4, 2006 #5

    VietDao29

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    ???
    Why on Earth are there still some people just post a complete solution (which is definitely, certainly, obviously, seriously, blah blah blah... against the forum's rules), without giving the OP an opportunity to solve the problem on his own??? :confused:
    Isn't your #2 post of this thread enough? Can't you wait for the OP to tell you if he can solve it or he still needs a little bit more help?
    Am I really missing something??? :confused:
     
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