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Integration problem

  1. Apr 7, 2007 #1
    The question is to evaluate the integral in the attachment. My idea was to use partial fraction decomposition and so my new integral looked like:

    A/((e^x)+1) + B/((e^x)+5) where I got A=-2 and B=-10

    When I integrate this, I get:

    -2ln((e^x)+1) - 10ln((e^x)+5)

    However, this is not the answer.
     

    Attached Files:

  2. jcsd
  3. Apr 7, 2007 #2

    Dick

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    I can tell you right away that the integral of 1/(e^x+1) is not ln(e^x+1). Differentiate the latter (remembering the chain rule) to see why not. To do this correctly substitute u=(e^x+1).
     
    Last edited: Apr 7, 2007
  4. Apr 7, 2007 #3
    I cannot see the integral till the attatchment it is approved but how did you manage to integrate something of the form

    [tex]\frac{A}{e^x+a}[/tex] to

    [tex]A\ln{(e^x+a)}[/tex]

    ?

    One way to check if your integration is done correctly is to differentiate the function you arrived at and see if it matches the earlier function.
     
  5. Apr 7, 2007 #4
    For ∫-2/((e^x)+1) dx, I choose u=(e^x)+1 so du=(e^x) dx so I get:

    -2 ∫ du / u(e^x) where (e^x)= u-1 therefore -2 ∫ du / (u^2)-u

    Is this correct?
     
  6. Apr 7, 2007 #5

    Dick

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    Yes. So you see you need another partial fractions thing to solve that. Because you WOULDN'T just write log(u^2-u), would you?
     
  7. Apr 7, 2007 #6
    No I would not!
     
  8. Apr 7, 2007 #7
    Got it... thanks a lot!
     
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