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Thread: 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 $\displaystyle u = (\ln 5x)^{2} $ and $\displaystyle dv = dx$

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

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

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

    so $\displaystyle \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; Mar 11th 2010 at 10:25 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.

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

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