$\int\limits_{\gamma} \frac{1}{z-1}$, $\gamma(\theta) = 2e^{i\theta}$, $0 \leq \theta \leq \frac{\pi}{2}$ $\gamma(\theta) = 2e^{i\theta}$ is a circle centered at $(0,0)$ with radius $2$, so $z = 1$ is inside this path and thus we have to use Cauchy's integral formula for $\int\limits_{\gamma} \frac{1}{z-1}$.
So Cauchy's integral formula says $\int\limits_{\gamma} \frac{f(z)}{(z-a)^{n+1}} = \frac{2\pi i}{n!} f^{n}(z)$.
If I choose $f(z) = 1$, then is this function still not holomorphic because the point $z = 1$ is on the path? If it is holomorphic, then the value of the integral is easy; it's just $2\pi i$. However, the value of the integral is $2\pi i$ for $0 \leq \theta \leq 2\pi$ over the entire path of the circle. How do I get the value for $0 \leq \theta \leq \frac{\pi}{2}$? Can I just divide $2\pi i$ by $4$ since $\frac{\pi}{2}$ is $\frac{1}{4}$ of the circle?
 A: For the first question: you can use Cauchy.  The formula applies to the integral
$$\int_C \frac{f(z)}{(z-a)^{n+1}}\,dz$$
where $f(z)$ is holomorphic on and inside $C$.  Note carefully: that's $f(z)$, not $f(z)/(z-a)^{n+1}$.  The latter is not holomorphic, except in trivial cases, but it doesn't matter.  Your $f(z)$ is just $1$, so it's holomorphic and you can use Cauchy.
Second question: if $\theta$ goes from $0$ to $\pi/2$ then you do not have a closed path and definitely cannot use Cauchy.  You need to parametrise the path, or use an appropriate antiderivative.  The integral will not be a quarter of the previous integral, except possibly by coincidence.
A: You need to go this way

$$ \int_{0}^{\pi/2} \frac{2i e^{i\theta} d\theta}{2e^{i\theta}-1}d\theta=\dots\,. $$

A: Reparametrize the curve as $$ \Gamma(t) = 1 + R(t) e^{it} \quad 0 \leq t \leq \pi- \arctan 4= \alpha $$ Note that, $R(t) = |2e^{it} -1|$, so no problem with continuity or differentiability wrt $t$. Now under this parametrization, $$ dz = (R'(t) + i R(t)) e^{it} dt , \qquad z-1 = R(t) e^{it}$$ So this leads to, $$ \int_\Gamma \frac{1} {z-1} dz = \int_0 ^{\alpha} (i + \frac{R'}{R})dt $$
$$\qquad  = \big[i t + \log R(t)\big]_0 ^{\alpha}=i(\pi -\arctan 4)+ \log \sqrt5 $$
