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How do you integrate (ln(x))^2? dx

  1. Jan 29, 2009 #1
    it seems you cant use the property ln x^n = n ln x.

    I'm thinking there's integration by parts involved but not sure.
     
  2. jcsd
  3. Jan 29, 2009 #2

    tiny-tim

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    Hi maxfails! :smile:

    Yes, use integration by parts with 1 as the function. :wink:
     
  4. Jan 29, 2009 #3
    ln xn = t
    x = et
    dx = et dt

    so initial eqn becomes

    [tex]\int t^n e^t dt[/tex]

    and now integrate by parts
     
  5. Jan 29, 2009 #4
    [tex]ln \ x^n = t[/tex]

    [tex]e^{ln \ x^n} = e^t[/tex]

    [tex]x^n=e^t[/tex]

    [tex]\frac{d}{dt} \ (x^n)=\frac{d}{dt} \ (e^t)[/tex]

    [tex]0=e^t[/tex]

    remember that:

    [tex]x=exp \ y \Leftrightarrow y=ln \ x[/tex]

    So

    [tex]0=e^t \Leftrightarrow t = ln \ 0[/tex]

    Since ln 0 is undefined, so t is undefined too...

    :confused:
     
  6. Jan 29, 2009 #5
    The problem is that this formula is [itex]\ln(x^n)=n\ln(x)[/itex], but you are now interested in [itex](\ln(x))^n[/itex] which is different. You probably knew this, but didn't sound very sure about it.

    Well tiny-tim of course answered quite sufficiently already, but I thought I would like to say that personally I like writing recursive formulas such as this:

    [tex]
    (\ln(x))^n = D_x \big(x(\ln(x))^n\big) - n(\ln(x))^{n-1}
    [/tex]

    fundoo, optics.tech, those were quite confusing comments :bugeye:
     
  7. Jan 29, 2009 #6
    Umm there's a difference between (ln(x))^2 and ln(x^2). The first is what you seem to have, the latter is 2*ln(x).
     
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