Here and here, we studied how to evaluate the contour integral $\oint_C f(z)dz$ when $f(z)$ is analytic everywhere within and on the positively oriented simple closed contour $C$ except for a finite number of isolated singularities interior to $C$. The calculation of residues however can be a pain if there are many isolated singularities of $f(z)$ interior to $C$. It turns out that by slightly modifying the function, we may just need to deal with only one isolated singularity regardless of how many isolated singularities of $f(z)$ there are interior to $C$. This gives a great advantage from computational viewpoint.

Theorem. If a function $f$ is analytic everywhere except for a finite number of isolated singularities interior to a positively oriented simple closed contour $C$, then
$$\oint_C f(z)dz=2\pi i\mathrm{Res}_{z=0}\left[\frac{1}{z^2}f\left(\frac{1}{z}\right)\right].$$

Proof.

From the above picture, we see that the function $f(z)$ has a Laurent series expansion
$$f(z)=\sum_{n=-\infty}^\infty c_n z_n\ (R_1<|z|<\infty)$$
where
$c_n=\frac{1}{2\pi}\oint_{C_0}\frac{f(z)}{z^{n+1}}dz$ $(n=0,\pm 1,\pm 2,\cdots)$. In particular, we have
$$\oint_{C_0}f(z)dz=2\pi ic_{-1}.$$
Since the condition of validity with the representation is not of the type $0<|z|<R_2$, $c_{-1}$ is not the residue of $f$ at $z=0$. Let us replace $z$ by $\frac{1}{z}$ in the representation. Then
$$\frac{1}{z^2}f\left(\frac{1}{z}\right)=\sum_{n=-\infty}^\infty\frac{c_n}{z^{n+2}}=\sum_{n=-\infty}^\infty\frac{c_{n-2}}{z^n}\ \left(0<|z|<\frac{1}{R_1}\right).$$
Hence,
$$c_{-1}=\mathrm{Res}_{z=0}\left[\frac{1}{z^2}f\left(\frac{1}{z}\right)\right]$$
and
$$\int_{C_0}f(z)dz=2\pi i\mathrm{Res}_{z=0}\left[\frac{1}{z^2}f\left(\frac{1}{z}\right)\right].$$
Since $f$ is analytic throughout the region bounded by $C$ and $C_0$ (topologically speaking $C_0$ is homotopic to $C$),
$$\oint_C f(z)dz=\oint_{C_0}f(z)dz.$$

Example. Evaluate $\oint_C\frac{5z-2}{z(z-1)}dz$ where $C:\ |z|=2$.

Solution. Let $f(z)=\frac{5z-2}{z(z-1)}$. Then
$$\begin{align*} \frac{1}{z^2}f\left(\frac{1}{z}\right)&=\frac{5-2z}{z(1-z)}\\ &=\frac{5-2z}{z}\cdot\frac{1}{1-z}\\ &=\left(\frac{5}{z}-2\right)(1+z+z^2+\cdots)\\ &=\frac{5}{z}+3+3z+\cdots\ (0<|z|<1). \end{align*}$$
Thus,
$$\oint_C\frac{5z-2}{z(z-1)}dz=2\pi i(5)=10\pi i.$$