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Logarithmic Integration

  1. Sep 21, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\int^{e^2}_{e}[/tex] [tex]\frac{1}{xlnx}[/tex] dx

    2. Relevant equations



    3. The attempt at a solution

    I substituted U= xlnx
    So DU= ([tex]\frac{1}{x}[/tex]dx..........so Du * X = 1dx

    From there I am stuck!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 21, 2009 #2

    rock.freak667

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    so if du =(1/x)dx where u=lnx


    what does

    [tex]\int \frac{1}{x lnx} dx[/tex]

    change to in terms of u and du?
     
  4. Sep 21, 2009 #3
    [tex]\frac{du}{u}[/tex] ?
     
  5. Sep 21, 2009 #4

    rock.freak667

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    yes so what is



    [tex]\int \frac{1}{u} du = ? [/tex]
     
  6. Sep 21, 2009 #5
    is it just LN (x)??
     
  7. Sep 21, 2009 #6

    rock.freak667

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    ln(u)

    now since you are integrating from e2 to e, what is your integral equal to in terms of x?
     
  8. Sep 21, 2009 #7
    ln e2- ln e
     
  9. Sep 21, 2009 #8

    rock.freak667

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    No.

    If you integrated 1/u du and got ln(u), and u = ln(x). What is ln(u) now?
     
  10. Sep 21, 2009 #9
    it is LN (LN(x))
     
  11. Sep 21, 2009 #10

    rock.freak667

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    right now so now compute Ln(ln(e2))-Ln(ln(e))
     
  12. Sep 22, 2009 #11
    alright so LN (1/e^2)- LN (1/e) ???
     
  13. Sep 22, 2009 #12

    VietDao29

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    No no, ln(ln(e2)) is definitely not ln(1/e2).

    Now, let's do it step by step then. What is ln(e2)?
     
  14. Sep 22, 2009 #13
    it is 2
     
  15. Sep 22, 2009 #14
    and LN(e) is 1, so it would be LN (2)- LN (1)? Which is .693??
     
  16. Sep 22, 2009 #15

    VietDao29

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    ln(1) = ln(e0) = 0, so, you can leave it as: ln(2) - ln(1) = ln(2). Taking an approximation is okay, though. :)
     
  17. Sep 22, 2009 #16
    ok so that's the final answer??
     
  18. Sep 22, 2009 #17

    VietDao29

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    Yup. :)
     
  19. Sep 22, 2009 #18
    Great! Thanks for the help everyone! I just didn't think it made sense to have an LN in another LN, but I guess that works!! Thanks again!
     
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