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This question is an exact duplicate of:

$\int \dfrac{w^2}{(w^2+1)^2(w^2+2w+2)}=\dfrac{p \pi}{q}$, limits are from minus infinity to infinity.

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marked as duplicate by user21820, YuiTo Cheng, postmortes, DMcMor, RRL Jun 26 at 15:20

This question was marked as an exact duplicate of an existing question.

  • $\begingroup$ Maple agrees with the locations of your poles. It agrees on the residue at $1\pm i$. But disagrees on your residue at $\pm i$. Once you have the residues, to find $p$ and $q$ you evaluate that integral using the residue theorem. Of course you need to invent an appropriate contour to use. $\endgroup$ – GEdgar Apr 21 '17 at 12:00
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    $\begingroup$ Use Cauchy Residue Theorem to evaluate the integral which should give you a rational number $p/q$ where $p$ and $q$ are integers. Hence, you have $p$ and $q$. You need the residues which you calculated to use the CRT... en.wikipedia.org/wiki/Residue_theorem Note, since the limits of the integral are $-\infty$ and $+\infty$ then you your path of integration will most likely be the half circle in the upper complex plane with infinite radius. $\endgroup$ – Pixel Apr 21 '17 at 12:19
  • $\begingroup$ For $z=i$:$$\lim_{z\to i}\frac{d}{dz}(z-i)^2\frac{z^2}{(z+i)^2(z-i)^2(z^2+2z+2)}$$ $\endgroup$ – Nosrati Apr 21 '17 at 12:30
  • $\begingroup$ Ok. I found your answer. $\endgroup$ – Nosrati Apr 21 '17 at 12:47
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    $\begingroup$ $\int f=2\pi i\sum Residues$ $\endgroup$ – Nosrati Apr 21 '17 at 12:57
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After more comments see \begin{eqnarray} residue(f,-1+ i) &=& \dfrac{3}{25}-\dfrac{4}{25}i\\ residue(f,-1- i) &=& \dfrac{3}{25}+\dfrac{4}{25}i\\ residue(f, i) &=& -\dfrac{3}{25}+\frac{9 i}{100}\\ residue(f, -i) &=& -\dfrac{3}{25}-\frac{9 i}{100} \end{eqnarray} for integral $\displaystyle\int_{-\infty}^\infty \dfrac{x^2}{(x^2+1)^2(x^2+2x+2)}dx$ which lies in upper half plane, only residues in $i$ and $-1+i$ are important that concludes $$2\pi i\left(\dfrac{3}{25}-\dfrac{4}{25}i-\dfrac{3}{25}+\frac{9 i}{100}\right)=2\pi i\left(-\frac{7 i}{100}\right)=\dfrac{7}{50}\pi$$

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  • $\begingroup$ Sorry, I didn't write more details. You'r welcome. $\endgroup$ – Nosrati Apr 21 '17 at 15:23
  • $\begingroup$ Sorry, which contour can I use in order to show that these residue's lie in that contour? $\endgroup$ – user407151 Apr 24 '17 at 11:42
  • $\begingroup$ Upper-half circle $|z|<R$ (in upper half plane) that $R\to\infty$. $\endgroup$ – Nosrati Apr 24 '17 at 11:48

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