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Let $\gamma(z_0,R)$ denote the circular contour $z_0+Re^{it}$ for $0\leq t\leq 2\pi$.
Evaluate $$ \int_{\gamma(0,1)}\dfrac{z^2+1}{z(z^2+4)}dz $$

I've tried to use the binomial expansion with $(z+\frac{1}{z})(z^2+4)^{-1}$ but then I'm not sure what to do. Any help will be great.

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$$f(z) = \frac{z^2+1}{z^2+4}$$ is analytic on the unit disc, so Cauchy's integral formula shows that the integral equals $2\pi i f(0) = \pi i/2$. (No need for the residue theorem)

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Use the Residue Theorem. The only pole within $\gamma(0,1)$ is at $z=0$. The residue there is 1/4, so the value of the integral is $i \pi/2$.

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