# Closed contour integral that contains singularities

Let $$f(z) = \frac{z^2+8z+42}{(z^2+4)(z^2-4z+7)}$$ I have that $$\lim_{R\to\infty} \int_{C_R} f(z)\ \text{d}z =0$$ where $C_R$ denotes the the circle with radius $R$, centre $0$, positively oriented.

I wish to show that $$\int_C f(z)\ \text{d}z =0$$ where $C$ is the circle with radius $5$, centre $2$, positively oriented.

Now $C$ contains all the singularities of $f$. How can I use the first result to show the above?

We show that $\int_{C_R}f=-\int_Cf$. To do this, take a PokeBall-like contour (or two separate contours, if you'd prefer), going from $R$ to $-R$ via the upper semicircle, from $-R$ to $-3$ via straight line, go one round about $C$ (positive orientation), $-3$ to $-R$ via straight line, then $-R$ to $R$ via bottom semicircle. This contour is a simple closed curve (note that if we did not take positive orientation about $C$ this would not be the case), and $f$ holomorphic with no singularities inside the region bounded by contour, so by Residue Theorem the integral is 0, but at the same time it is $\int_{C_R}f+\int_Cf$.