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Math Help - Integration By Parts

  1. #1
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    Integration By Parts

    How to integrate x * radical(2x+1)?

    it looks like this: wolfram site

    the problem says to do it using tabular integration by parts, which is what I did, but I ended up with this:
    x^3/3*(2x+1)^(3/2) - 3x^2/15*(2x+1)^(5/2)+6x/105*(2x+1)^(7/2)+6/945*(2x+1)^(9/2)
    I know that's really hard to see.. so here's the image: wolfram (the first box, where it says input)

    it's so different from the actual answer in my calculator and wolfram! does anybody know how to integrate it properly?
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  2. #2
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    \displaystyle u=x^3 \ \ du=3x^2dx \ \ \ dv=\sqrt{2x+1} \ \ v=\frac{(2x+1)^{3/2}}{3}

    \displaystyle uv-\int vdu
    Last edited by dwsmith; December 23rd 2010 at 08:17 PM.
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  3. #3
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    wait.. if dv=\sqrt{2x+1}, shouldnt v=(2x+1)^{3/2}/3?
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  4. #4
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    Yup.
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  5. #5
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    hm.. but the answer I got after doing that was

    \displaystyle \frac{x^3}{3} (2x+1)^{3/2} - \frac{3x^2}{15} (2x+1)^{5/2} +\frac{6x}{105} (2x+1)^{7/2}-\frac{6}{945} (2x+1)^{9/2}+constant

    when according to my calculator and wolframalpha.com, it should be

    \displaystyle \frac{x^3}{9} (2x+1)^{3/2} - \frac{x^2}{21} (2x+1)^{3/2} +\frac{2}{105} (2x+1)^{3/2} - \frac{2}{315} (2x+1)^{3/2} +constant
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  6. #6
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    \displaystyle u=x^3 \ \ du=3x^2dx \ \ \ dv=\sqrt{2x+1} \ \ v=\frac{(2x+1)^{3/2}}{3}

    \displaystyle uv-\int vdu

    \displaystyle \frac{x^3(2x+1)^{3/2}}{3}-\int x^2(2x+1)^{3/2}dx

    \displaystyle s=x^2 \ \ ds=2xdx \ \ \ dt=(2x+1)^{3/2} \ \ t=\frac{(2x+1)^{5/2}}{5}

    \displaystyle uv-\left[st-\int tds\right]

    \displaystyle \frac{x^3(2x+1)^{3/2}}{3}-\frac{x^2(2x+1)^{5/2}}{5}+\int \frac{2x(2x+1)^{5/2}}{5}dt

    \displaystyle w=2x \ \ dw=2dx \ \ \ dz=\frac{(2x+1)^{5/2}}{5} \ \ z=\frac{(2x+1)^{7/2}}{35}

    \displaystyle uv-st+\left[wz-\int zdw\right]

    \displaystyle \frac{x^3(2x+1)^{3/2}}{3}-\frac{x^2(2x+1)^{5/2}}{5}+\frac{2x(2x+1)^{7/2}}{35}-\int \frac{2(2x+1)^{7/2}}{35}dx

    Is this the expression you solved?
    Last edited by dwsmith; December 23rd 2010 at 09:04 PM. Reason: Changed equation to expression
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  7. #7
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    huh. um.. well actually I did the first part of it after your first post, but it looked like what I had initially tried and got, so I stopped.

    the directions in the book actually told me to find the integral using tabular integration by parts, which is something like this:



    and so I tried something similar, listing under the column "u"
    x^3
    3x^2
    6x
    6
    0

    and under the column "dV"
    \(2x+1)^{1/2}
    \frac{(2x+1)^{3/2}}{3}
    \frac{(2x+1)^{5/2}}{15}
    \frac{(2x+1)^{7/2}}{105}
    \frac{(2x+1)^{9/2}}{945}

    then I just connected all of them and got the result that I got. I'm not sure where I went wrong..
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  8. #8
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    If we continue from here

    \displaystyle \frac{x^3(2x+1)^{3/2}}{3}-\frac{x^2(2x+1)^{5/2}}{5}+\frac{2x(2x+1)^{7/2}}{35}-\int \frac{2(2x+1)^{7/2}}{35}dx,

    we will obtain:

    \displaystyle \frac{x^3(2x+1)^{3/2}}{3}-\frac{x^2(2x+1)^{5/2}}{5}+\frac{2x(2x+1)^{7/2}}{35}-\frac{2(2x+1)^{9/2}}{315}+C

    \displaystyle \frac{(2x+1)^{3/2}}{315}\left[105x^3-63x^2(2x+1)+18x(2x+1)^2-2(2x+1)^3\right]+C

    And continue simplifying.

    Here is Mathematica 7's answer

    Last edited by dwsmith; December 23rd 2010 at 09:01 PM. Reason: Forgot my +C
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