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Integration by parts

  1. May 10, 2007 #1
    intergrate (ln(x))^2

    so i set u=(lnx)^2...which makes du=2lnx(1/x)

    then i set dv=dx...which makes v=x

    according to the formula for integration by parts i have

    x(lnx)^2- integral x(2lnx)(1/x)
    simplifying it i get x(ln)^2-2intergral lnx

    and here is where i am stuck....what i the integral of lnx?
  2. jcsd
  3. May 10, 2007 #2


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    The derivative of (lnx)^2=(2lnx)/x.

    [edit: of course that's you've written... i glanced and say ln(1/x)... sorry :blushing: ]

    A hint for integrating lnx; use parts, taking dv=dx and u=lnx
    Last edited: May 10, 2007
  4. May 10, 2007 #3
    How about integration-by-parts once again? :)
  5. May 10, 2007 #4


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    xln(x) - x looks good from where I'm standing.

    I just wondered "what function gives ln(x) when differentiated? Well ln(x)' = 1/x. So what if I try xln(x)? Now I get ln(x) + 1. So I need to add something to the mix that gives -1 when differentiated." Hence xln(x) - x.
  6. May 10, 2007 #5
    ohhh yes......do integration by part again........

    thank you!
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