Simple question in verifying Cauchy's integral theorem I'm trying to verify Cauchy's integral theorem "by hand" for
$$
I = \oint_C\frac{z}{2z-5}dz\,,
$$
where $C$ is the circle $\vert z-3\vert=2$. From Cauchy's integral theorem, the answer is $5\pi i /2$ although I'm not quite sure how to finish the last step of my verification.
My attempt is as follows.
\begin{equation*}
\begin{aligned}
I &= \oint_C \frac{z-3}{2(z-3)+1}dz+\oint_C\frac{3}{2(z-3)+1}dz \\
&= \int_0^{2\pi i} \frac{e^{2x}}{e^x+1/4}dx + \frac{3}{2}\int_0^{2\pi i} \frac{e^{x}}{e^{x}+1/4}dx \\
&= \int_0^{2\pi i} \frac{e^x(e^x+1/4-1/4)}{e^x+1/4}dx + \frac{3}{2}\left[\log{\left(e^{ix}+1/4\right)}\right]_0^{2\pi i} \\
&= \left[e^{x} + \frac{5}{4}\log{\left(e^{x}+1/4\right)}\right]_0^{2\pi i} \\
&= \frac{5}{4}\log{\left(e^x+1/4\right)}\bigg\vert_0^{2\pi i}
\end{aligned}
\end{equation*}
How do I show that the evaluation of the log term is $2\pi i$? The factor of $1/4$ in the $\log$ is throwing me off. I suspect that I am to use
$$
\log z = \log r + i(\theta+2\pi n)
$$
here, although if I am, I feel silly because I'm not seeing a simple way to employ it.
 A: Write $I = \oint_C\frac{z}{2z-5}dz=\frac12\oint_C \frac{z-\frac52+\frac52}{z-\frac52}dz=\frac12 \underbrace{\oint_Cdz}_{=0}+\frac54 \oint_C\frac{1}{z-\frac52}dz$. Hence, the problem really amounts to solving $$II=\frac54\oint_C \frac{1}{z-\frac52}dz,$$
which can be done only using elementary methods. The trick in the picture below, we see that the integral along $C_1+C_2$ is equivalent to the integral along $C$ plus the integral along the clockwise circle, centered at $z=\frac52$ of radius, say, $r$.
Hence, $\oint_{C_1}+\oint_{C_2}=\oint_C-\oint_{C_r}$, where $C_r$ is the positively oriented counterclockwise circle. Rearranging, we see that
$$
II=\frac54 \oint_{C_1} \frac{1}{z-\frac52}dz+\frac54\oint_{C_2} \frac{1}{z-\frac52}dz+\frac54\oint_{C_r} \frac{1}{z-\frac52}dz.
$$
Now the key for the first two integrals is that since we are avoiding the singularity, there exists an analytic logarithm such that  $\frac{d}{dz}\log(z-\frac52)=\frac{1}{z-\frac52}$. Hence, by the fundamental theorem, the integral around a closed loop is zero. This leaves the last integral, which is easy to solve with parameterization, since the circle is centered at $z=\frac52$.

