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Math Help - integration by partial fractions

  1. #1
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    integration by partial fractions

    i need to take the integral of: x/(x^2 + 4 * x + 13) using partial fractions. my teacher told me to start off by turning this into the integral of: x/(x + 2)^2 +9, but i have no idea what to do when i get it in that form, anyone care to shed some light on this for me?
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  2. #2
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    Krizalid's Avatar
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    \int {\frac{x}<br />
{{x^2  + 4x + 13}}\,dx}  = \frac{1}<br />
{2}\int {\frac{{2x}}<br />
{{x^2  + 4x + 13}}\,dx}. So

    \int {\frac{x}<br />
{{x^2  + 4x + 13}}\,dx}  = \frac{1}<br />
{2}\left( {\int {\frac{{2x + 4}}<br />
{{x^2  + 4x + 13}}\,dx}  - \int {\frac{4}<br />
{{(x + 2)^2  + 9}}\,dx} } \right).

    The first integral is immediate, now take a look to the second one, it's an arctangent, can you do that?
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  3. #3
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    ok, so i i have been trying to work through this and now i am at:

    integral of x/(x+2)^2 +9 , i see that if i can set u = to x+2 i can use the arcsin thing, however i have that stupid x on top that won't allow me to use that substitution, can anyone help me out?
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  4. #4
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    Krizalid's method is probably the best for this integral, but if you want to practise integrating after partial fractions (which is very useful) then the following may be useful:
    u = x+2 => x = u-2
    <br />
\frac {u-2}{u^2} = \frac {u}{u^2} - \frac {2}{u^2}

    Good job finding the right substitution to use.
    Last edited by badgerigar; January 31st 2008 at 04:56 PM. Reason: fixed formatting
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  5. #5
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    Ok here is how I would go about this:

    \int \frac{x}{(x + 2)^2 + 9} dx = \int \frac{x}{9((\frac{x + 2}{3})^2 + 1)} dx = \tfrac{1}{9} \int \frac{x}{(\frac{x + 2}{3})^2 + 1} dx

    \tan \theta = \frac{x+2}{3}
    \sec^2 \theta d \theta = \tfrac{1}{3} dx


    = \tfrac{1}{9} \int \frac{3\overbrace{(3\tan \theta - 2)}^{x} \sec^2 \theta}{(\tan \theta)^2 + 1} d \theta = \tfrac{1}{3} \int \frac{(3 \tan \theta - 2) \sec^2 \theta}{\sec^2 \theta} d \theta = \tfrac{1}{3} \int (3 \tan \theta - 2) d \theta


    = \tfrac{1}{3} \Big[ 3 \int \tan \theta d \theta - 2 \theta \Big] = \tfrac{1}{3} \Big[ 3 \int \tan \theta d \theta - 2 \theta \Big]  = - \tfrac{1}{3} \Big[3ln|\cos (\tfrac{x + 2}{3})|  + 2 \arctan (\tfrac{x + 2}{3}) \Big]


    = \boxed{- \tfrac{1}{3} \Big[3 ln|\tfrac{3}{\sqrt{x^2 + 4x + 13}}|  + 2 \arctan (\tfrac{x + 2}{3}) \Big]}

    I probably made a mistake somewhere but that seems like the general idea.
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