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Tricky integral

  1. Jun 30, 2008 #1
    1. The problem statement, all variables and given/known data
    1/ [e^x (1-e^(-2x))^1/2]

    2. Relevant equations
    integration by parts

    3. The attempt at a solution
    First I split up the integral so it would be (1/e^-x)[(1/(1-e^-2x)^1/2]
    Then I set u=e^-x, dv= (1-e^-2x)^-1/2
    du= -e^-x
    For my v I got 2(1-e^-2x)^1/2 but I don't know how to go about with the e^-2x, as in how to finish solving for the v-part.

    I tried u-substitution and I get another term in the denominator. Then when I tried to differentiate the answer I originally got for v, I did not have anything in the denominator... can anyone help?
  2. jcsd
  3. Jun 30, 2008 #2


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    Homework Helper

    [tex]\int \frac{1}{e^x \sqrt{1-e^{-2x}}} dx[/tex]

    try [itex]t^2=1-e^{-2x}[/itex]
  4. Jun 30, 2008 #3
    Is that valid? I just tried to solve it using your suggestion and with u=e^-x. With the u-subst I end up getting the arcsin integral, but I get something different with the t^2 substitution.
  5. Jun 30, 2008 #4
    That's what I did too and verified it by taking it's derivative, but Mathematica gives a totally different solution.

    I changed mine to:



    [tex]u=e^{-x} \rightarrow \ln{u}=-x[/tex]
  6. Jun 30, 2008 #5
    Oh interesting, why did you take the natural log? Where does that come into play?
  7. Jun 30, 2008 #6
    I let:

    [tex]u=e^{-x}[/tex] be my u-sub

    [tex]-\frac{du}{u}=dx[/tex] be my du-sub
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