Complex line integral $\int \frac{1}{2z-1} dz$ around the unit circle. Consider the complex integral
$$\int_{C} \frac{1}{2z-1}dz$$
where $C$ is the unit circle oriented counter-clockwise. I know that since the integrand is not analytic inside the unit circle (particularly $z = 1/2$) the integral over the closed curve is not necessarily zero. I have a theorem which says

Let $C$ be a piecewise smooth path, represented by $z=z(t)$, where $a \leq t \leq b$. Let $f(z)$ be a continuous function on $C$. Then
$$\int_{C} f(z) dz = \int_{a}^{b}f[z(t)]z'(t)dt.$$

Thus choosing the paramterization $z(t) = e^{it}$ for $0\leq t \leq 2\pi$ with $z'(t) = ie^{it}$, I get the integral
$$ I = \int_{0}^{2\pi}\frac{ie^{it}}{2e^{it}-1}dt = \frac{1}{2}\ln(2e^{it}-1) \bigg|_{0}^{2\pi}.$$
Now since the principle branch of the logarithm is $\text{Ln} \ z:= \ln|z|+i \text{Arg} \ z$ where $-\pi <  \text{Arg} z \leq \pi$ and $\ln \ z := \text{Ln} \ z +i2\pi k$ for $k\in \mathbb{Z}$
we get
$$ \frac{1}{2}\ln(2e^{it}-1) \bigg|_{0}^{2\pi} = \frac{1}{2}[\text{Ln}(2e^{i2\pi}-1)+ i2\pi] - \frac{1}{2}\text{Ln}(2e^{0}-1)=i\pi.$$
Question: Is this method of evaluation correct? I am mostly uncertain if the last step pertaining to $\text{Ln}$ is correct.
 A: Your proof is correct. I can only add a tiny detail. You define the principal logarithm as $\operatorname{Ln} \ z:= \ln|z|+i \text{Arg} \ z$. Then you continue to define the "general" complex logarithm as $\ln \ z := \operatorname{Ln} \ z +i2\pi k$. As you can see this contradicts your previous definition of the principal logarithm since you define $\ln$ in two different ways. To prevent this I would refer to the general complex logarithm as $L$, so $$L(z) := \operatorname{Ln} \ z +i2\pi k.$$

In the comment section you wrote that you have not yet reached the residue theorem. This post will provide a shorter proof with the help of Cauchy's integral formula, a special case of the residue theorem.

Let $U$ be an open subset of the complex plane $\mathbb{C}$, and suppose the closed disk $D$ defined as $D = \bigl\{z:|z - z_0| \leq r\bigr\}$ is completely contained in $U$. Let $f\colon U \to \mathbb{C}$ be a holomorphic function, and let $γ$ be the circle, oriented counterclockwise, forming the boundary of $D$. Then for every $a$ in the interior of $D$, $$f(a) = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z-a}\,\mathrm{d}z.$$

Now for your case, let $\gamma=C=\{z\in\mathbb{C}:\lvert z\rvert =1\}$, $f(z)=1/2$ and $a=1/2$. Putting all of this together,
$$\frac{1}{2}=\frac{1}{2\pi i}\oint_{\lvert z\rvert =1} \frac{1}{2(z-\frac{1}{2})}\,\mathrm{d}z=\frac{1}{2\pi i}\oint_{\lvert z\rvert =1} \frac{1}{2z-1}\,\mathrm{d}z,$$
which means that
$$\oint_{\lvert z\rvert =1} \frac{1}{2z-1}\,\mathrm{d}z=\frac{2\pi i}{2}=\pi i.$$
