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Integrals involving 'e'

  1. Sep 6, 2008 #1
    1. The problem statement, all variables and given/known data
    So, I have to integrate the following expression:
    (e^7x)/(e^(14x) + 9)dx


    2. Relevant equations

    We are doing the section on 'integrating by u-substitution' right now, so that might help in finding a solution...



    3. The attempt at a solution
    So, I tried a bunch of stuff - I tried u=7x, u=e^7x, u=14x, u=e^14x, u=x, etc. etc. and I can't get the right answer!!! I have been working on this problem for over an hour and am on the verge of tears!
     
  2. jcsd
  3. Sep 6, 2008 #2

    cristo

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    Hint: e14x=(e7x)2
     
  4. Sep 6, 2008 #3
    I know that.. but how do I work from there???
    The problem with "e" is that its derivative is different than integrating something like x^2
     
  5. Sep 6, 2008 #4

    cristo

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    But you can use a subsitution: use, say, u=e^{7x}
     
  6. Sep 6, 2008 #5
    How??
    I tried that million times and I couldn't get the right answer!!
    Here's what I tried

    u=e^7x
    du=(7e^7x)dx

    so then the integrand (that's what the thing being integrated is called, right?)
    becomes
    (u/(u^2 + 9))*du/(7e^7x)

    that doesn't work!!!!!
     
  7. Sep 6, 2008 #6

    cristo

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    Can you simplify the line in bold (perhaps using (#)?)

    And yes, the thing being integrated is called the integrand :smile:
     
  8. Sep 6, 2008 #7
    sorry for my constant refrain of complaining, but I already tried that and couldn't figure out how to integrate (u^2 + 9)^-1........

    I know it involves something with ln(u^2 + 9), but I couldn't get that to work either!
     
  9. Sep 6, 2008 #8

    cristo

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    Oh, ok, well I can see why you're banging your head against a brick wall then! Integrals like [tex]\int\frac{dx}{1+x^2}[/tex] are ones you should really know (especially for exams etc..) but in essence most people just look them up. The way to solve this, however, would be to make a final substitution x=tany.
     
  10. Sep 6, 2008 #9
    Wait, what???
    I don't follow - I have never seen that before in Calculus I (I'm in Calc. II right now)

    And what about the 9??
    That derivative is for x^2 + 1, not x^2 + 9
     
  11. Sep 6, 2008 #10
    I mean that integral instead of derivative
     
  12. Sep 6, 2008 #11
    Oh waiiiiiit never mind I found it in my reference page (integral of du/(x^2 + a^2) = blahblah.

    Cristo, thank you so much - you are amaaaaaaaaaazing and although I did cry over this problem for about ninety minutes, I have learned a bunch and I am going to memorize some of these integrals involving u^2-a^2 and u^2+a^2, etc. that we didn't cover in AP Calc with flashcards.

    Thanks again!!
     
  13. Sep 6, 2008 #12

    cristo

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    Glad I could help.

    In case you're interested, the way to transform x^2+9 into the above form is to make a substitution of the form v=x/3, which will give 9(v^2-1). Then use v=tany So, you can solve this problem using around 4 different substitutions in succession.
     
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