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Integration Question

  1. Jan 3, 2008 #1
    Greetings -

    This integral is part of a larger problem I'm working on - I'm stuck (I Think)

    Here's the integral:

    [tex] = \int -\frac{e^{-x}}{e^{-3x}(1+e^x)}dx[/tex]

    I've tried some algebra and substitutions:

    [tex] = \int -\frac{e^{2x}}{(1+e^x)}dx[/tex]

    [tex] = u=e^x[/tex]
    [tex] = du=e^x dx[/tex]

    [tex] = \int -\frac{u}{(1+u)}du[/tex]

    Is this correct? Is this a standard form for something? - looking like ln()?

    OR
    could go:
    [tex] = \int -\frac{e^{-x}}{e^{-3x}(1+e^x)}dx[/tex]

    [tex] = \int -\frac{1}{e^{-2x} +e^{-x}}dx[/tex]
     
    Last edited: Jan 4, 2008
  2. jcsd
  3. Jan 3, 2008 #2
    Split it in two partial fractions

    [tex]\frac{u}{1+u}=\frac{u+1-1}{1+u}=1-\frac{1}{1+u}[/tex]
     
    Last edited: Jan 3, 2008
  4. Jan 4, 2008 #3
    Thanks - I'm still trying to hack my way around to correctly post the problem - I'm trying to find examples of this [tex] stuff, I can't quite get the problem to post correctly.
     
  5. Jan 4, 2008 #4
  6. Jan 4, 2008 #5
    So my integral is
    [tex] \int du -\int \frac{1}{1+u} du[/tex]

    I know there will be a ln() involved eventually
    ?
     
  7. Jan 4, 2008 #6
    Yes!
    Which is equal to what?
     
  8. Jan 4, 2008 #7
    Here's what I'm getting - can you confirm?

    [tex] -e^x + ln|1+e^x| [/tex]
     
    Last edited: Jan 4, 2008
  9. Jan 4, 2008 #8
    Correct!

    Only you forgot an overall minus sign, one from the inital integral!
     
  10. Jan 4, 2008 #9
    oops

    thanks

    By the way, can you suggest any other method that would have solved this integral?

    I have not used partial fractions in a long time, if you had not suggested it, I would not have gone there.

    The integral looked simple enough for me to "pencil and paper" it and impress my friends without the need for a CRC handbook or other aid - I was wrong.
     
  11. Jan 4, 2008 #10
    But this is the more efficient way! To try to transform integrals with log's, sin's, cos's, exp's to rational form, and then apply partial fractions!

    At least that' s the way I think! :smile:
     
  12. Jan 4, 2008 #11
    Well you could have used polynomial division if you had not thought of adding +/- 1 to the numerator so that you could split it into two.
     
  13. Jan 4, 2008 #12
    I guess I need to knock off the rust and practice up a little bit so the next time it's more obvious.

    Thank you all for helping with this -
     
  14. Jan 12, 2008 #13
    Instead of using partial fractions. Just use another substitution. Use say t=1+u. This implies that u=t-1. So you have integral of (t-1)/t. And keep in mind that dt=du so this integral is relatively easy. Because its just the integral of 1-(1/t). which is t-ln(t). Just back sub for the t. so 1+u-ln(1+u)
     
    Last edited: Jan 12, 2008
  15. Jan 12, 2008 #14

    Gib Z

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    Homework Helper

    You used partial fractions.
     
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