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The derivative of (ln(x))^2

  1. Oct 8, 2009 #1
    Is this a u-sub? I went through and got 1/(2x)^2 but I am not sure if that is correct.
     
  2. jcsd
  3. Oct 8, 2009 #2

    danago

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    You can use a substitution, but the answer is not quite 1/(2x)^2.

    How did you arrive at that answer?
     
  4. Oct 8, 2009 #3
    Actually I asked the wrong question.

    I'm working on an integration by parts problem it asks me to integrate what is in the topic title.
    I went about it by saying that U = ln(x)^2 and that dv = 1.

    When I went through and plugged everything into the integration by parts formula I arrived at

    xln(x)-(integral sign)(1)(1/(2x)^2)

    The thing that I am getting hooked up is trying to take that integral of 1/(2x)^2. I think I messed up somewhere but I am not sure where to look. Any suggestions?
     
  5. Oct 8, 2009 #4

    danago

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    Hmm im not really sure how you ended up xln(x)-(integral sign)(1)(1/(2x)^2).

    I would choose the same parts as you did, i.e. u = (ln x)^2 and dv = 1, but im not sure you applied the formula correctly.

    [tex]\int u dv =uv - \int v du[/tex]

    Is that what you are using?
     
  6. Oct 8, 2009 #5
    Yeah that's what I am using. du = 1/(2x^2) v= x

    (lnx)^2)(x)-(integral)(1/(2x)^2)(1)

    Did I take the derivative of (ln(x))^2 wrong?
     
  7. Oct 8, 2009 #6

    danago

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    Yea maybe have another look at the derivative of (ln x)2

    You could use the substitution z = ln(x), or treat it as (ln x)2 = (ln x)(ln x) and apply the product rule.
     
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