I've been asked to compute
$\displaystyle\int_{0}^{\infty}\frac{x^2}{(x^2+1)^2(x^2+2x+2)}dx$
via finding the poles of the integrand and then construncting a contour to take advantage of cauchy's residue theorem. I have already computed the residues, but I keep coming up with something like $\frac{5 \pi}{3}$. However it is now to my realization that the integral is not symmetric about 0, so we have to find a different approach...
Thanks! EDIT: this isn't homework, I've been doing some integration problems on my own to brush up on it.
This may be done by considering the integral
$$\oint_C dz \frac{z^2 \log{z}}{(z^2+1)^2 (z^2+2 z+2)}$$
where $C$ is a keyhole contour of large radius $R$ and small radius $\epsilon$ about the positive real axis, as pictured below.
If you work out the integrals along each part of the contour $C$, we get that this contour integral is equal to
$$\int_{\epsilon}^R dx \frac{x^2 \log{x}}{(x^2+1)^2 (x^2+2 x+2)} + i R \int_0^{2 \pi} d\theta \, e^{i \theta} \frac{R^2 e^{i 2 \theta} \left (\log{R} + i \theta \right)}{(R^2 e^{i 2 \theta}+1)^2 (R^2 e^{i 2 \theta} + 2 R e^{i \theta} + 2) } \\ + \int_R^{\epsilon} dx \frac{x^2 (\log{x}+i 2 \pi) }{(x^2+1)^2 (x^2+2 x+2)}+ i \epsilon \int_{2 \pi}^0 d\phi\, e^{i \phi} \frac{\epsilon^2 e^{i 2 \phi} \left (\log{\epsilon} + i \phi \right)}{(\epsilon^2 e^{i 2 \phi}+1)^2 (\epsilon^2 e^{i 2 \phi}+ 2 \epsilon e^{i \phi}+2)} $$
We take the limit as $R \to \infty$ and $\epsilon \to 0$; the second and fourth integrals vanish and the contour integral reduces to
$$-i 2 \pi \int_0^{\infty} dx \frac{x^2}{(x^2+1)^2 (x^2+2 x+2)} $$
By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the sum of the residues at the poles of the integrand of the contour integral. For this integrand, there are double poles at $z=\pm i$ and simple poles at $z=-1 \pm i$.
Now, before we compute the residues, it is imperative that we get the branch of the logarithm correct. In this case, we assumed that the branch is taken about the positive real axis, so that the arguments of the poles must run between $0$ and $2 \pi$. Thus, the poles are taken to be at (double poles) $z=e^{i \pi/2}$, $z=e^{i 3 \pi/2}$, (simple poles) $z=\sqrt{2} e^{i 3 \pi/4}$, $z=\sqrt{2} e^{i 5 \pi/4}$.
For the residue calculation, I will simply state the results without showing the differentiation and algebra, which I think the reader should be able to do on his/her own.
$$\begin{align}\operatorname*{Res}_{z=e^{i \pi/2}} \frac{z^2 \log{z}}{(z^2+1)^2 (z^2+2 z+2)} &= \left [\frac{d}{dz} \frac{z^2 \log{z}}{(z+i)^2 (z^2+2 z+2)} \right ]_{z=e^{i \pi/2}}\\ &= \left (-\frac{1}{10}-i \frac{1}{20} \right ) - \left (\frac{9}{200}+ i \frac{3}{50} \right ) \pi\end{align}$$
$$\begin{align}\operatorname*{Res}_{z=e^{i 3 \pi/2}} \frac{z^2 \log{z}}{(z^2+1)^2 (z^2+2 z+2)} &= \left [\frac{d}{dz} \frac{z^2 \log{z}}{(z-i)^2 (z^2+2 z+2)} \right ]_{z=e^{i 3 \pi/2}}\\ &= \left (-\frac{1}{10}+i \frac{1}{20} \right ) + \left (\frac{27}{200}- i \frac{9}{50} \right ) \pi\end{align}$$
$$\begin{align}\operatorname*{Res}_{z=\sqrt{2} e^{i 3 \pi/4}} \frac{z^2 \log{z}}{(z^2+1)^2 (z^2+2 z+2)} &= \left [ \frac{z^2 \log{z}}{(z^2+1)^2 (2 z+2)} \right ]_{z=\sqrt{2} e^{i 3 \pi/4}}\\ &= \left (\frac{3 \log{2}}{50}-i \frac{2 \log{2}}{25} \right ) + \left (\frac{3}{25}+ i \frac{9}{100} \right ) \pi\end{align}$$
$$\begin{align}\operatorname*{Res}_{z=\sqrt{2} e^{i 5 \pi/4}} \frac{z^2 \log{z}}{(z^2+1)^2 (z^2+2 z+2)} &= \left [ \frac{z^2 \log{z}}{(z^2+1)^2 (2 z+2)} \right ]_{z=\sqrt{2} e^{i 5 \pi/4}}\\ &= \left (\frac{3 \log{2}}{50}+i \frac{2 \log{2}}{25} \right ) + \left (-\frac{1}{5}+ i \frac{3}{20} \right ) \pi\end{align}$$
We form the sum of these residues; from the above analysis, we conclude that the integral we seek is the negative of this sum. Of course, the imaginary part of the sum of the residues vanishes, as you will verify. The final result is
$$\int_0^{\infty} dx \frac{x^2}{(x^2+1)^2 (x^2+2 x+2)} = \frac15 - \frac{3 \log{2}}{25} - \frac{\pi}{100} \approx 0.0854$$
This was verified in Mathematica.