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Homework Help: Homework: Indefinite Integral

  1. May 3, 2008 #1
    Evaluate the indefinite integral:


    -I think that u=e^(2x)
    so then du=e^(2x)dx
    then the answer would end up being [(e^(4x)+9)/(-1)]^(-1)

    but it was incorrect; I think that my u might be wrong and that's where the problem is, but I am not sure. Please help, thank you!
  2. jcsd
  3. May 3, 2008 #2
    as in

    [tex]\int (\frac{e^{4x}}{e^{8x}}+9)dx ?[/tex]
    [tex]\int (e^{4x-8x}+9)dx = - \frac{1}{4}e^{-4x}+9x[/tex]

    or as

    [tex]\int (\frac{e^{4x}}{e^{8x}+9})dx ?[/tex]
    and then i don't know how to solve this =D
    If that's what you were asking, you got a tough one.
    [still thinking]

    It looks like an integration by parts question, or I am sleepy and can't see the answer D=
    But integration by parts doesn't work in my case still...
    Last edited: May 3, 2008
  4. May 3, 2008 #3
    correction on problem

    the second part is the one that we need help on...thank you!!
  5. May 3, 2008 #4


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    \frac{1}{4} \int(\frac{1}{u^{2}+3^{2}})du

    Can you finish it from there?
    Last edited: May 3, 2008
  6. May 3, 2008 #5
    As I was saying.. I must've been smoking something....
  7. May 3, 2008 #6
    I always get confused when taking antiderivatives of fractions...how do I go about doing that?
  8. May 3, 2008 #7
    natural log?
  9. May 3, 2008 #8
    So, I ended up with 1/4lnabs((e^4x)^(2))+9) and got it wrong, what am I doing incorrectly?
  10. May 3, 2008 #9


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