Integral of $\int_C \frac{1}{z^2-z} dz$ where $C$ is a circle of radius $2$ I wish to evaluate
$$ \int_C \frac{1}{z^2-z} dz \ \ \ \ \text{where} \ \ C=\left \{ z \in \mathbb{C}:| z |=2\right \}$$
I use the parameterisation $ \ \  \ \ \ \gamma(t)=2e^{it} \ \ : t=0\rightarrow 2\pi$
And so:
$$ \int_C \frac{1}{z^2-z} dz =\int_0^{2\pi}\frac{2ie^{it}}{4e^{2it}-2e^{it}}dt=\int_0^{2\pi}\frac{ie^{-it}}{2-e^{-it}}dt=\left [\log (2-e^{-it}) \right]_0^{2\pi} =0$$
What argument can I make in terms of closed curves? 
I know that the integral of an analytic function on a closed curve is $0$ but how can I show that $\frac{1}{z^2-z}$ is analytic on $C$ ?
 A: The function $f(z) = \frac{1}{z^2 - z}$ is not analytic on $\mathbb{C}$, and is not even defined at $z = 0$ or $z = 1$. Where it is defined, we have the partial fraction decomposition
$$ f(z) = \frac{1}{z(z-1)} = \frac{1}{z-1} - \frac{1}{z}$$
and so for the closed curve of radius 2 centred at the origin, the residue theorem shows that the integral is $0$, since the $\frac{1}{z-1}$ term will contribute $2 \pi i$ to the integral, and the $\frac{-1}{z}$ term will contribute $-2\pi i$.
If you were to take a circle of radius $\frac{1}{2}$, you should find that the integral will be $2 \pi i$, showing that $f(z)$ cannot be analytic on $\mathbb{C}$.
A: $z^2-z = z(z-1) = 0$ if $z=0$ or $z=1,$ so $\dfrac 1 {z^2-z}$ is $\infty$ when $z=0$ or $z=1.$
Now notice that your curve $C$ encloses both $0$ and $1.$
It is hazardous to talk about $\log$ when you're dealing with complex numbers rather than positive real numbers. With complex numbers $\log$ is at best a "multiple-valued function."
But $z\mapsto\dfrac 1 {z^2-z}$ is analytic everywhere in $\mathbb C$ except at $z=0$ and $z=1,$ so its integral along any curve that winds once around those two points is just the sum of its integrals around any curve that winds once around each of those points. So let $C_0$ and $C_1$ be circles of some small radius $r$ winding counterclockwise around $0$ and $1$ respectively.
\begin{align}
\int_{C_1} \frac{dz}{z(z-1)} & = \left( (\text{the value of }\frac 1 z \text{ at } z = 1) \times \int_{C_1} \frac{dz}{z-1} \right) \\[10pt]
& = 1\times \int_0^{2\pi} \frac{rie^{it}\,dt}{re^{it}} = 2\pi i.
\end{align}
The other one is dealt with similarly.
A: $f(z)=\frac{1}{z^2-z}$ is a meromorphic function with simple poles at $z=0$ and $z=1$.
Since $|f(z)|=O\left(\frac{1}{|z|^2}\right)$ as $|z|\to +\infty$, the given integral equals zero. 
This happens since the residue theorem ensures that, for any $R>1$ 
$$ I(R) = \oint_{|z|=R}f(z)\,dz $$
is constant, equal to $2\pi i\,\left(\text{Res}_{z=0}+\text{Res}_{z=1}\right) f(z)$. On the other hand, by the triangle inequality
$I(R) = O\left(\frac{1}{R}\right)$ as $R\to +\infty$, hence $I(2)=0$.
