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Another U substitution

  1. Sep 10, 2007 #1
    I know this must be similar.....

    [tex]\int \frac{e^x}{1+e^{2x}}[/tex]

    should [tex]u=1+e^{2x}[/tex]?

    Casey
     
  2. jcsd
  3. Sep 11, 2007 #2
    actually this one's a bit trickier.

    if you try 1+e^(2x) you won't really go anywhere since the derivative of e^(2x) is 2e^(2x) >< which doesn't appear on the top of the fraction.

    So is there any other substitution you can try? one that when derived will give you the quantity that is on top of the fraciton? o.o
     
  4. Sep 11, 2007 #3
    I see something, bit I can't figure out how to word it......I know [tex]e^{2x}=e^{(x)2}[/tex]...or something......
     
  5. Sep 11, 2007 #4
    yes o.o so what other substitution can you try? 1+e^(2x) doesn't work so..

    ps.
    (this time you won't be able to do the entire quantity on the bottom of the fraction).
     
  6. Sep 11, 2007 #5
    [tex]e^x[/tex] is the only thing that keeps popping into my head...but..
     
  7. Sep 11, 2007 #6
    but? o.o why don't you try it?

    when doing integrals it's not wierd to try one method for like 1/2 a page then say "this is going nowhere" and going some other way >< from the start....
     
  8. Sep 11, 2007 #7
    is this calc 2? b/c all it is the integral of arctan after you u subst. for e^x.
     
  9. Sep 11, 2007 #8
    I am.....and I am getting....hey wait is this a true statement? [tex](e^x)^2=e^{2x}[/tex]

    Casey
     
  10. Sep 11, 2007 #9
    yes.

    [tex]u=e^{x}[/tex]

    [tex]du=e^xdx[/tex]

    what is the integral of arctan?
     
  11. Sep 11, 2007 #10
    Ah! It's late....I had myself convinced that [tex]e^{2x}[/tex] and [tex](e^x)^2[/tex] were not equal....silly algebra.

    Thanks again and goodnight,
    Casey
     
  12. Sep 11, 2007 #11

    HallsofIvy

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    I think you mean "what is the derivative of arctan!:rolleyes:
     
  13. Sep 11, 2007 #12
    argh! yes actually that would be correct, lol.
     
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