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Integral of (1/x)(lnx)^2

  1. May 17, 2004 #1
    I don't know what i'm doing wrong when taking this simple integral.

    The integral is:

    [tex] \int[/tex] (1/x)(lnx)^2 dx

    [tex] \int[/tex] (2/x)(lnx) dx

    2[tex] \int[/tex] (lnx/x) dx

    Let u = lnx
    du/dx = 1/x
    dx = xdu

    2[tex] \int[/tex] (u/x) xdu

    2[tex] \int[/tex] (u) du

    2[tex] \int[/tex] (u^2)/2 + C

    (2(lnx)^2)/2 + C

    (lnx)^2 + C

    The answer is obviously wrong....how do i solve this properly?
  2. jcsd
  3. May 17, 2004 #2
    Try the substitution u = lnx.

  4. May 17, 2004 #3


    User Avatar
    Science Advisor

    How did you make that step? ln(x2)= 2ln(x)
    but this is (ln(x))2.

    Just go ahead and make the substitution u= ln(x) right at the start.
  5. May 19, 2004 #4

    Thanks..seems to work now. I got it. Thanks for all your help.
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