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Math Help - trouble with integral by parts

  1. #1
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    trouble with integral by parts

    I'm trying to solve an integral by parts and got stuck. I'm at step 4 of my work.

    I THINK I've got it right but it could be because of how I set up the whole thing that could be throwing me off.
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  2. #2
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    Sorry, are you referring to actual integration by parts, cause I don't see the need to use it.

    Your original equation can be simplified to  \displaystyle \int \frac{x^3}{(x+19)^2} ~dx

    Long division should give you
    <br />
\displaystyle \int -38 ~dx +  \displaystyle \int x ~ dx -6859  \displaystyle \int \frac{1}{(x+19)^2} ~dx + 1083 \displaystyle \int \frac{1}{x+19} ~dx

    The first one is integrating a constant. You should be able to do that...

    The second one is integrating a single x. That should be easy too

    The third one, you can integrate it directly if you're comfortable, or simply use the substitution u = x + 19 (remember the limits change during substitution, so turn the u back into x after integrating or change your limits)

    The fourth one integrates into a log.
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  3. #3
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    <br />
\displaystyle{\int \frac {x^3}{(x+19)^2}dx}=<br />

    <br />
\displaystyle{\int \frac {( \; (x+19)-19)^3}{(x+19)^2}dx}<br />

    Use this formula

    <br />
(a+b)^3=a^3+3a^2b+3ab^2+b^3<br />

    where
    <br />
a=x+19 \; \; <br />
b=-19<br />
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  4. #4
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    Here's a way to avoid long division:

    You have successfully written it as \displaystyle \int \frac{x^3}{(x+19)^2}\;{dx}.

    Let y = x+19, then x^3 = (y-19)^3 and dy = dx , so we have:

    \displaystyle \int \frac{x^3}{(x+19)^2}\;{dx} = \int \frac{(y-19)^3}{y^2}\;{dy}

    The rest is routine. Have you tried learning the Latex?
    It might make the task of posting and getting help easier.
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