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Transcendental Functions Containing Natural Logs

  1. Sep 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Evaluate the integrals

    a) y= eqn40561.png
    b) y= eqn4056.png

    2. Relevant equations

    Untitled-1.png

    3. The attempt at a solution

    I've tried using the Reciprocal Rule and substitution of each ln function with u to get a workable result, but havent come anywhere near a recognizable solution. The answers stated in the book are as follows:
    a) 1/ln4
    b) (ln2)^2

    [STRIKE]This tells me that, somehow a must end up looking like this:
    ln 1 - ln 4
    and b must look like:
    ln2 * ln2[/STRIKE]

    These would seem counter intuitive to my understanding of how these kinds of problems are solved and Im at a loss of how to fix it.
     
    Last edited: Sep 5, 2012
  2. jcsd
  3. Sep 5, 2012 #2

    LCKurtz

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    Science Advisor
    Homework Helper
    Gold Member

    You want the anti-derivative (definite integral), not the derivative, and there is no ##y## in either problem. In both problems try the substitution ##u=\ln x##.
     
  4. Sep 5, 2012 #3

    Mark44

    Staff: Mentor

    1/ln(4) ≠ ln(1) - ln(4), so it would be good for you to review the properties of logs.

    The expression on the right can be simplified like so:

    ln(1) - ln(4) = 0 - ln(4) = ln(1/4)

    Note the ln(1/4) is a negative number, while 1/ln(4) is positive, so they couldn't possibly be equal.
     
  5. Sep 5, 2012 #4
    I see the intent in setting up the integral as ##u*du## and anti-deriving to ##2u## but I dont see how ##du## could possibly equal ##1/x*dx## the derivative of ##\ln x## is ##(1/x)(d/dx*x)## which leaves us with ##u*du*dx## right?

    I also updated the errors in my original post. Thanks for catching them, I am definitely on the tail end of burned out.
     
  6. Sep 5, 2012 #5

    Mark44

    Staff: Mentor

    No. If u = lnx, then du = d(lnx) = d/dx(lnx)*dx = 1/x * dx
     
  7. Sep 5, 2012 #6

    HallsofIvy

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    Ouch! The anti-derivative of u is [itex](1/2)u^2+ C[/itex]

    Your original integral was
    [tex]\int \frac{dx}{x ln(x)}[/tex]
    (the second "dx" must be a typo.)

    If you let u= ln(x) then [itex]du= \frac{1}{x}dx= \frac{dx}{x}[/itex] so the integral becomes
    [tex]\int\frac{dx}{x ln(x)}= \int \frac{1}{ln(x)}\frac{dx}{x}= \int \frac{1}{u}du[/tex]

    "the derivative of ##\ln x## is ##(1/x)(d/dx*x)## "
    I can't make sense of that! The derivative of ln(x) is 1/x. Perhaps you mean (1/x) times the derivative of x with respect to x. That last would be 1 so that is the same as 1/x. Having written du/dx= 1/x, in differential terms, du= (1/x)dx.
     
  8. Sep 7, 2012 #7
    Thanks guys, despite the errors on my part, you suggestions and clarifications were awesome and I got it worked out :biggrin:
     
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