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Integral of a Natural Log

  1. Feb 5, 2004 #1

    It has been over a year since I last took calculus. And I don't recall how to take the integral of a natural logarithmic function. Here is the question that I am supposed to integrate.

    double integral 1/(x+y) dA


    R = [1,2] X [0,1]

    So what I did first was integrate with respect to y first. I ended up with


    with an upper limit of 1 and a lower limit of 0. Once simplified I get

    ln(x+1) - ln x or ln( (x+1)/x )

    Now I have to integrate with respect to x. But I can't remember how to take the integral of a natural log function. How do I proceed from here?

    I can't remember if I can do the following:

    Let G(x) = integral of ln( (x+1)/x ) dx


    e^G(x) = integral of e^ln( (x+1)/x ) dx

    which would simplify to

    integral of (x+1)/x dx.

    After I get a solution to the above equation I would then take the log of

    ln e^G(x) = ln (answer)

    to get

    G(x) = ln (answer).

    Can I do that? I can't remember. If not, how do I proceed from here?

    Any help is appreciated. Thankyou.
  2. jcsd
  3. Feb 5, 2004 #2


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    You can't move the exponential into the integral like that.

    That said, IIRC:

    [tex]\int ln(x) dx= x ln(x)-x[/tex]
    (You can derive this by using parts)
  4. Feb 5, 2004 #3

    I should have known I could have done it by parts. I see that now. Thanks NateTG.
  5. Feb 6, 2004 #4


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    And it's a lot easier to integrate [itex]\ln(x+1) - \ln x[/itex] than it is to integrate [itex]\ln (x+1)/x[/itex]. :smile:
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