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Integrals of e.

  1. Sep 6, 2007 #1
    I need to find the integral of (1/(1+e^x)) dx.

    ...I don't think I can use u substitution, but well, I'm basically open to suggestions.
     
  2. jcsd
  3. Sep 6, 2007 #2
    It should be solvable with substitution. Hint: u = 1+e^x
     
  4. Sep 6, 2007 #3
    but, then wouldn't du = e^x dx? That would require a (1/e^x) outside the integral to balance it out, but my Cal I teacher (I'm now in Cal II with a different teacher) told me that I couldn't do this.
     
  5. Sep 6, 2007 #4
    Write x in terms of u, and then take the differential.
     
  6. Sep 6, 2007 #5
    let u = 1 + e^x
    then we know du = (e^x)dx
    and e^x = u -1

    is there something we can multiply both the numerator and the denominator by to make this substitution work? Hint: this will turn it into a partial fraction decomposition problem.
     
  7. Sep 6, 2007 #6
    Multiply by 1. (1=e-x/e-x)
     
  8. Sep 6, 2007 #7

    HallsofIvy

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    That multiplication isn't really necessary. JonF's first point was sufficient: if u= 1+ ex, then ex= u- 1. du= exdx= (u-1)dx so dx= du/(u-1).
    [tex]\int\frac{dx}{1+e^x}= \int\frac{du}{(u-1)(u)}[/tex]
    which can be done by partial fractions.
     
    Last edited: Sep 6, 2007
  9. Sep 6, 2007 #8
    Since i'm clumsy at partial fractions, I find multiplying by 1 to be a bit easier.
     
  10. Sep 6, 2007 #9
    Edit: woops sorry just saw d_leet mentioned this method already. But if you have a fraction with some e^x or e^-x then if you cant do u substitution, multiply by (e^-x)/(e^-x) or (e^x)/(e^x) and then do a nice u substitution.

    another method:

    multiply the top and bottom by e^-x to get:

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

    which you can use the substitution u=e^-x +1 then you have du = e^-x dx at the top which reduces the problem down to a simple integral
     
    Last edited: Sep 6, 2007
  11. Sep 7, 2007 #10

    VietDao29

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    Multiplying both numerator, and denominator by ex, or, making a substitution, is pretty much the same. After that, you have to do Partial Fraction it. You cannot avoid it. :)
     
  12. Sep 7, 2007 #11
    I could waste time by going the trig-way. :wink:
     
  13. Sep 7, 2007 #12
    Really? I ended up with du/u which doesn't seem like I need to do partial fractions.
     
  14. Sep 8, 2007 #13
    Then I’m pretty sure you did something wrong, what substitution did you use?
     
  15. Sep 8, 2007 #14
    I used e^-x + 1 for u then du= -e^-x dx so since that is on the top already with just a different constant you can get -du/u.
     
  16. Sep 8, 2007 #15
    It’s the integral of: (1/(1+e^x)), not (1/(1+e^-x)). That "-" makes a huge difference. If you use that substitution you can’t substitute the bottom.
     
  17. Sep 8, 2007 #16

    HallsofIvy

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    Nor was the integrand
    [tex]\frac{e^{-x}dx}{1+ e^{-x}}[/tex]
    as PowerIso seems to think!
     
  18. Sep 8, 2007 #17
    no what PowerIso did was this I think:

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

    which doing the substitution u=e^(-x)+1 gives you -du/u.
     
  19. Sep 8, 2007 #18
    I dont even think partial fractions are necessary. u(u-1)=u^2-u

    so

    [tex]\int\frac{dx}{1+e^x}= \int\frac{du}{(u-1)(u)} = \int{1 \over u^2-u}du = ln(u^2-u) + C = ln(e^{2x} + e^x) + C[/tex]
     
  20. Sep 8, 2007 #19
    Well, here is what I did. I multiplied by e^-x/e^-x and I got e^-x/(e^-x + 1) , so i'm still fairly certain it works.
     
  21. Sep 8, 2007 #20
    Yes that is what I did :).
     
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