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The integral is the following:

$$\int_{|z|=r} \frac{z+1}{z(z^2+4)} dz , r>0, r \neq 2 $$

I'm a little bit lost, I know that its partial fraction expansion is

$$ \frac{z+1}{z(z^2+4)} = \frac{4-z}{4(z^2+4)} + \frac{1}{4z} $$ but this doesn't seem like a Laurent series expansion and I don't know from here what is supposed to be the residue, so I can calculate the integral by residues.

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    $\begingroup$ As a short cut for the case $r > 2$. Notice the function is analytic for $|z| > 2$ and the modulus of the integrand falls off faster than $r^{-1}$ as $r \to \infty$. You can evaluate the integral by deform the contour to infinity and conclude the integral is $0$ when $r > 2$. $\endgroup$ – achille hui Dec 4 '14 at 13:30
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Put

$$f(z)=\frac{z+1}{z(z^2+4)}$$

so

$$\begin{align*}&\bullet\;\;\;\text{Res}_{z=0}(f)=\lim_{z\to 0}zf(z)=\frac14\\ &\bullet\;\;\;\text{Res}_{z=\pm 2i}(f)=\lim_{z\to\pm2i}(z\pm2i)f(z)=\frac{1\pm2i}{\pm2i(4i)}=\pm\frac18(1\pm2i)\end{align*}$$ $${}$$

Now use the Residue theorem for integrals, depending on whether $\;0<r<2\;$ , or $\;r>2\;$

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    $\begingroup$ I think your last residues should be $(1\pm2i)/(-8)$. $\endgroup$ – David Dec 4 '14 at 13:35
  • $\begingroup$ So if $0<r<2$ we have that the integral around $|z|=r$ is the sum of the residues whose of the poles norm is less than 2 (times $2 \pi i$), which only would be 1/4 when 0<r<2. So the result in this case is $\frac{\pi i}{2}$. And the r>2, the integral equals $2 \pi i (\frac{1}{4} + (\frac{1}{8} + \frac{i}{4}) + (\frac{-1}{8} + \frac{-i}{4})) = \frac {\pi i}{2}$ $\endgroup$ – Di Alejandra Dec 4 '14 at 13:39
  • $\begingroup$ @David, there was a typo in the last quantity, but for that I think it is right now. $\endgroup$ – Timbuc Dec 4 '14 at 13:43
  • $\begingroup$ Almost so, @Alejandra: for $\;r>2\;$ all the simple poles have to be taken into account. Yet check those residues carefully: I could have made a mistake. $\endgroup$ – Timbuc Dec 4 '14 at 13:43

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