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Integrartion by parts

  1. Oct 1, 2006 #1
    How do you evaluate [tex] \frac{1}{x\ln x} [/tex] by integration by parts. I tried doing this doing the following:

    [tex] u = \frac{1}{\ln x}, du = \frac{-1}{x(\ln x)^{2}}, dv = \frac{1}{x}, v = \ln x [/tex]. So I get:

    [tex] \int udv = uv-\int vdu = 1 + \int \frac{1}{x\ln x} = \int \frac{1}{x\ln x} [/tex]. I know the answer is [tex] \ln(\ln x) [/tex]

    Last edited: Oct 1, 2006
  2. jcsd
  3. Oct 1, 2006 #2
    hmm.....its not so complicated dude.....the ans is ln(lnx). Ur f(x)=(lnx)^-1 and ur f'(x) is x^-1 . get it?
  4. Oct 1, 2006 #3
    ok i got it
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