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Simple Integration

  1. Oct 25, 2005 #1
    Hi. Im trying to integrate (e^x +1)/(e^x -1)
    I have looked at this for almost an hour and don´t know how to start with it. I want to use substitution but I have to rewrite this in some way. Could anyone please give me a little hint?
     
  2. jcsd
  3. Oct 25, 2005 #2

    Physics Monkey

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    Swatch, try breaking the integral into two pieces and using a different substitution on each piece. One of the pieces will immediately look easy, and the other will be easy after one line of algebra.
     
  4. Oct 25, 2005 #3

    HallsofIvy

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    Perhaps I'm missing something but a kind of obvious substitution would be
    u= ex. Then (ex+1)/(ex-1) becomes (u-1)/(u+ 1). Of course, du= exdx so dx= (1/u)du. The integral
    [tex]\int \frac{e^x+1}{e^x-1}dx[/tex] becomes
    [tex]\int \frac{u+1}{u(u-1)}du[/tex]
    which can be done with partial fractions.
    If you don't like partial fractions, let u= ex-1. Then du= exdx so dx= (1/u) du again but now ex+ 1= u+ 2. The integral becomes
    [tex]\int\frac{u+2}{u^2}du= \int \left(u^{-1}+ 2u^{-2}\right)du[/tex].
     
    Last edited: Oct 25, 2005
  5. Oct 26, 2005 #4
    I integrated with u=e^x and used partial fraction
    to get
    -ln(e^x) + 2*ln(e^x -1) +C

    I differentiated this back to the beginning, so I should be right. But I got a different looking answer in my textbook.
     
  6. Oct 26, 2005 #5

    TD

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    If the derivative of your solution gives the integrand again, you should be OK.
    It's possible to find different forms of answers when integrating, that very much depends on the method used.
     
  7. Oct 26, 2005 #6
    thank you for your help
     
  8. Oct 26, 2005 #7
    mathmatica says it is

    -2x+xln(x)+Li(x)
     
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