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Math Help - Integration of parts

  1. #1
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    Integration of parts

    Stuck on a question using integration of parts.
    integral sign x^2ln4x dx
    u=ln4x v=1/3x^3
    du= dv=x^2

    S u dv = uv - S v du
    ln4x( )-S1/2x^3( )
    I don't know how to find du above? Is it 1/x(4x)??

    Any help would be great, thanks.
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  2. #2
    Member anonimnystefy's Avatar
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    Re: Integration of parts

    hi bradycat

    you can look at it like this:
    if u=ln(4x) then the derivative of u with respect to x is
    du/dx=4/4x=1/x
    multiply both sides by dx and you get:
    du=dx/x

    also dv should be:
    v=1/3 x^3
    dv/dx=x^2
    dv=x^2 dx.

    don't forget your dx.
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  3. #3
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    e^(i*pi)'s Avatar
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    Re: Integration of parts

    u = \ln(4x) \Leftrightarrow u' = \dfrac{1}{x}

    x^2 \ln(4x) - \int \left(\dfrac{1}{2}x^3 \times \dfrac{1}{x}\right)\ dx

    = x^2 \ln(4x) - \dfrac{1}{2} \int x^2 \ dx
    Last edited by e^(i*pi); August 14th 2011 at 12:17 PM. Reason: TYPO in the first line
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  4. #4
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    Re: Integration of parts

    so in my answer
    (1/3)x^3ln4x-(1/9)x^3+c equals

    1/3(X^3) ((ln4x-1/3))=c??
    Last edited by bradycat; August 14th 2011 at 01:20 PM.
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  5. #5
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    e^(i*pi)'s Avatar
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    Re: Integration of parts

    Quote Originally Posted by bradycat View Post
    so in my answer
    1/3X^3(ln4x-1/3)=c??
    No, x^3 isn't a common factor. and you've put \dfrac{x^3}{9} when it should be \dfrac{x^3}{6} in the second term.

    What do you get after integrating?
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  6. #6
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    Re: Integration of parts

    You choose to differentiate x\mapsto \ln 4x and integrate x\mapsto x^2.

    \int x^2 \ln 4x \textrm{d}x = \frac{x^3}{3}\ln 4x - \int \frac{x^3}{3}\cdot \frac{1}{x} \textrm{d}x = \frac{x^3}{3}\ln 4x -\frac{1}{3} \int  x^2 \textrm{d}x

    So :

    \int x^2 \ln 4x \textrm{d}x = \frac{x^3^}{3}\big(\ln 4x - \frac{1}{3}\big) + \lambda
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  7. #7
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    Re: Integration of parts

    Quote Originally Posted by e^(i*pi) View Post
    No, x^3 isn't a common factor. and you've put \dfrac{x^3}{9} when it should be \dfrac{x^3}{6} in the second term.

    What do you get after integrating?
    Slight misunderstanding, the OP had already integrated. (But = c wants to be + c of course.)

    As confirmed by Hugal. But here's a pic, just in case it helps...



    ... where (key in spoiler) ...

    Spoiler:

    ... is the product rule, straight continuous lines differentiating downwards with respect to x.

    ... is lazy integration by parts, doing without u and v.


    __________________________________________________ __________
    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
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  8. #8
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    Re: Integration of parts

    thank you everyone. That is the answer i got above, I just forgot to put in my brackets to seperate the 1/3 and then the x to the power of 3.
    THANK YOU. for clarifying for me. Have a great day
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  9. #9
    Member anonimnystefy's Avatar
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    Re: Integration of parts

    you're welcome.have a good one as well.
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