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Integral with e^x in it

  1. Sep 6, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]\int\frac{1+e^{x}}{1-e^{x}}[/tex] dx


    2. Relevant equations
    .


    3. The attempt at a solution
    I've tried substituting for u=e^x or u=1-e^x but I can't seem to get anywhere. Haven't done calc in a while and just want someone to point me in the right direction. Thanks.
     
    Last edited: Sep 6, 2009
  2. jcsd
  3. Sep 6, 2009 #2
    Letting [itex] u = e^x [/tex] gives
    [tex]
    \int\frac{1+u}{1-u}\frac{1}{u}\,du
    [/tex]
    Next expand in partial fractions. The integrand becomes
    [tex]
    \frac{2}{1-u} + \frac{1}{u}
    [/tex]
    which you can easily integrate.
     
    Last edited: Sep 7, 2009
  4. Sep 6, 2009 #3
    I'm thinking [tex]u=e^{x}[/tex] might work, at least for x<0.

    Also, [tex]{{1+u}\over{1-u}}={{1-u+2u}\over{1-u}}=1+2{{u}\over{1-u}}[/tex]
     
    Last edited: Sep 6, 2009
  5. Sep 6, 2009 #4
    For x>0, you might try [tex]u=e^{-x}[/tex]

    [tex]{{1+e^{x}}\over{1-e^{x}}}={{e^{-x}+1}\over{e^{-x}-1}}={{u+1}\over{u-1}}={{2u}\over{u-1}}-{{u-1}\over{u-1}}[/tex]



    [tex]\int({1\over{u}}-{2\over{u-1}})du[/tex]
     
    Last edited: Sep 6, 2009
  6. Sep 7, 2009 #5

    VietDao29

    User Avatar
    Homework Helper

    Well, why can't you sub u = ex, when x > 0?
     
  7. Sep 7, 2009 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You can- it really doesn't make any difference.
     
  8. Sep 7, 2009 #7
    Thanks a lot guys.
     
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