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Math Help - Problem with integration by parts

  1. #1
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    Problem with integration by parts

    Hey guys/gals,

    I have a problem involving integration by parts that has me stumped. Here is the equation:

    \int {xln(4+x^2)}dx

    Problem is, I assume that the u will be ln(4+x^2) and the dv will be xdx, but when I solve for that, my equation ends up being:

    \frac {\1}{2}x^2ln(4+x^2) - \int \frac {x^3}{4+x^2}dx

    ...which may be right, but I don't know what my next step should be. Any help on this would be greatly appreciated, just having a total brainfart with this one.


    Edit: only took 50 tries to get both equations formatted right... you know you're a noob when... :P
    Last edited by billb91; February 4th 2013 at 10:27 PM.
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  2. #2
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    Re: Problem with integration by parts

    I suggest, you divide the numberator with the denominator... using division algorithm (manual division that is to say), then after you have separated the fraction.. integrate .. you cannot decompose the function since the numerator has a higher degree.. also.. you cannot use inverse tangent..
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  3. #3
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    Re: Problem with integration by parts

    Quote Originally Posted by billb91 View Post
    Hey guys/gals,

    I have a problem involving integration by parts that has me stumped. Here is the equation:

    \int {xln(4+x^2)}dx

    Problem is, I assume that the u will be ln(4+x^2) and the dv will be xdx, but when I solve for that, my equation ends up being:

    \frac {\1}{2}x^2ln(4+x^2) - \int \frac {x^3}{4+x^2}dx

    ...which may be right, but I don't know what my next step should be. Any help on this would be greatly appreciated, just having a total brainfart with this one.


    Edit: only took 50 tries to get both equations formatted right... you know you're a noob when... :P
    This should not be done with Integration by Parts, at least not to begin with...

    \displaystyle \begin{align*} \int{x\ln{\left( 4 + x^2 \right) }\,dx} &= \frac{1}{2}\int{2x\ln{\left( 4 + x^2 \right) } \, dx} \end{align*}

    Now let \displaystyle \begin{align*} w = 4 + x^2 \implies dw = 2x\,dx \end{align*} and the integral becomes

    \displaystyle \begin{align*} \frac{1}{2}\int{2x\ln{\left( 4 + x^2 \right)}\,dx} &= \frac{1}{2} \int{\ln{(w)}\,dw} \end{align*}

    Now using integration by parts with \displaystyle \begin{align*} u = \ln{(w)} \implies du = \frac{1}{w}\,dw \end{align*} and \displaystyle \begin{align*} dv = 1\,dw \implies v = w \end{align*} we have

    \displaystyle \begin{align*} \frac{1}{2}\int{\ln{(w)}\,dw} &= \frac{1}{2}\left[ w\ln{(w)} - \int{w\left( \frac{1}{w}\right) dw} \right] \\ &= \frac{1}{2} \left[ w\ln{(w)} - \int{1\,dw} \right] \\ &= \frac{1}{2} \left[ w\ln{(w)} - w \right] + C \\ &= \frac{1}{2}w \left[ \ln{(w)} - 1 \right] + C \\ &= \frac{1}{2}\left( 4 + x^2 \right) \left[ \ln{ \left( 4 + x^2 \right) } - 1 \right] + C \end{align*}
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    Re: Problem with integration by parts

    Quote Originally Posted by billb91 View Post
    Edit: only took 50 tries to get both equations formatted right... you know you're a noob when... :P
    And thank you for giving us something that's easy to read...

    One suggestion: instead of "ln", you can use "\ln". It adds a little space and un-italicizes it:

    \int {xln(4+x^2)}dx

    \int {x\ln(4+x^2)}dx

    - Hollywood
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  5. #5
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    Re: Problem with integration by parts

    Thanks so much for the help guys! Setting it up the way prove it did made things way, way simpler!
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