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Math Help - Integration problem with ln (using parts)

  1. #1
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    Integration problem with ln (using parts)




    I've never worked with this kind of problem before. Can anyone please provide a step-by-step answer?
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  2. #2
    Super Member Random Variable's Avatar
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    let  u = (\ln 5x)^{2} and  dv = dx

    then  du = 2 \ln (5x)*\frac{5}{x} \ dx and  v =x

     \int (ln 5x)^{2} = x (\ln 5x)^{2} - 10 \int  \ln 5x \ dx

    Integrating by parts again, you can show that  \int \ln 5x = \frac{1}{5}x \ln 5x -\frac{1}{5}x +C

    so  \int (ln 5x)^{2} = x (\ln 5x)^{2} - 2x \ln 5x + 2x + C
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  3. #3
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    How were you able to get du?
    Last edited by Archduke01; March 11th 2010 at 11:25 AM.
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  4. #4
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Archduke01 View Post
    How were you able to get du?
    I used the chain rule.

     \frac{du}{dx} = 2*(\ln 5x)^{1}*\frac{d}{dx} \  \ln 5x = 2(\ln 5x) * \frac{5}{x}

    so  du = 2(\ln 5x) * \frac{5}{x} \ dx
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  5. #5
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    If \frac {d}{dx} ln 5x = 5/x then how did you get (1/5)x in the 2nd integration by parts?
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