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Intergrating 1/[xlog(x)]

  1. Nov 26, 2004 #1
    This is doing my head in!

    I split it to 1/x * 1/log(x) and got the intergral = 1 + the intergral when using intergration by parts. :cry:

    I know the answer is log[log(x)] but have no idea how you get log of a log.

    Got a feeling the answer is going to really obvious!
  2. jcsd
  3. Nov 26, 2004 #2


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    Substitute u=log(x)
  4. Nov 26, 2004 #3
    log(x) = u, u=exp(x), x du = dx.
    intergral becomes exp[-u] 1/u exp du
    = 1/u du
    = log u
    = log[log(x)]

    Cheers! :biggrin:
  5. Nov 27, 2004 #4
    \int{frac{1}{x\ln{x}}}\d x=\int{frac{1}{\ln{x}}}\d \ln{x}
  6. Nov 27, 2004 #5
    [tex]\int{frac{1}{x\ln{x}}}\d x=\int{frac{1}{\ln{x}}}\d \ln{x}[\tex]
  7. Nov 27, 2004 #6
    [tex]\int{frac{1}{x\ln{x}}}\d x=\int{frac{1}{\ln{x}}}\d \ln{x}[/tex]
  8. Nov 27, 2004 #7


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    Do you mean?


    You can click on the above for the code that generated it.

    You know you can preview posts before you submit. You can also edit and delete!
  9. Nov 27, 2004 #8
    I see. Sorry for making a big mess here.
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