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Math Help - partial fractions (calculus)

  1. #1
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    partial fractions (calculus)

    integral of (x-3)/(x^2 + 2x + 4)
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  2. #2
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    Quote Originally Posted by twilightstr View Post
    integral of (x-3)/(x^2 + 2x + 4)
    \int \frac{x-3}{x^2+2x+4}dx = \int \frac{x-3}{(x+1)^2+3}dx If u = x+1 then du = dx and we get

    \int \frac{u-4}{u^2+3}\,du. Split the integral up and do each separately.
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  3. #3
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    Since the denominator is not easily factorable and PFD does not help much, we can try completing the square.

    \int\frac{x-3}{x^{2}+2x+4}dx

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

    Now, let u=x+1, \;\ du=dx, \;\ x=u-1

    \int\frac{u-4}{u^{2}+3}du=\int\frac{u}{u^{2}+3}du-4\int\frac{1}{u^{2}+3}du

    Now, it's just a matter of ln and arctan.
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  4. #4
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    how would you evaluate the integral if the denominator was squared: (x^2 + 2x + 4)^2
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  5. #5
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    On post #2 was told that x^2+2x+4=(x+1)^2+3, now you have to compute \int{\frac{dx}{{{\big({{\left( x+1 \right)}^{2}}+3\big)}^{2}}}}.

    Put \sqrt3u=x+1 and the integral becomes \frac{1}{3\sqrt{3}}\int{\frac{du}{{{\left( {{u}^{2}}+1 \right)}^{2}}}}. Now, for this one, put u=\frac1v and the integral becomes,

    \begin{aligned}<br />
   -\int{\frac{{{v}^{2}}}{{{\left( {{v}^{2}}+1 \right)}^{2}}}\,dv}&=-\int{\left\{ -\frac{1}{2\left( {{v}^{2}}+1 \right)} \right\}'v\,dv} \\ <br />
 & =-\left( -\frac{v}{2\left( {{v}^{2}}+1 \right)}+\frac{1}{2}\int{\frac{dv}{{{v}^{2}}+1}} \right) \\ <br />
 & =\frac{1}{2}\left( \frac{v}{{{v}^{2}}+1}-\arctan v \right)+{{k}_{1}}.<br />
\end{aligned}

    Now put everything together and we get,

    \int\frac{dx}{\big(x^2+2x+4\big)^2}\,dx=\frac{1}{6  \sqrt{3}}\left( \frac{\sqrt{3}(x+1)}{{{x}^{2}}+2x+4}+\arctan \frac{\sqrt{3}}{x+1} \right)+k.
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