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Math Help - [SOLVED] Integration by parts with a natural log

  1. #1
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    [SOLVED] Integration by parts with a natural log

    \int 2xln(3x)~dx

    I set:
    u=ln(3x)
    dv=2x dx
    v=x^2
    du=1/x

    I came up with:
    <br />
x^2ln(3x) - \int x^2 * 1/x~dx<br />

    But I can't figure out what to do with the part under the integral, because after multiplying I am left with \int x~dx. I am sure I messed up somewhere.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by redman223 View Post
    \int 2xln(3x)~dx

    I set:
    u=ln(3x)
    dv=2x dx
    v=x^2
    du=1/x

    I came up with:
    <br />
x^2ln(3x) - \int x^2 * 1/x~dx<br />

    But I can't figure out what to do with the part under the integral, because after multiplying I am left with \int x~dx. I am sure I messed up somewhere.
    nope, you're correct

    now finish up
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  3. #3
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    you are ok ... \int x \, dx = \frac{x^2}{2} + C

    you finally get x^2\ln(3x) - \frac{x^2}{2} + C

    taking the derivative ...

    x^2 \cdot \frac{1}{x} + 2x \cdot \ln(3x) - x = 2x \cdot \ln(3x)<br />

    you're back where you started.
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  4. #4
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    Wow, I can't believe that I didn't see that earlier, thanks. I guess my brain is taking a little while longer to come off of summer break. It's amazing how much can be forgotten in as little as 6 weeks.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by redman223 View Post
    Wow, I can't believe that I didn't see that earlier, thanks. I guess my brain is taking a little while longer to come off of summer break. It's amazing how much can be forgotten in as little as 6 weeks.
    then you know how bad i have it! my summer break was more than twice that long!
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