antiderivative of $\frac{1}{z(z-1)}$, complex logarithm I have the domain $\mathbb{C} \backslash [0,1]$ and want to show that $$\int_\gamma \frac{1}{z(z-1)}dz = 0$$
for all closed curves $\gamma$. I want to accomplish this by explicitly finding an antiderivative.
I did a partial fraction decomposition and got
$$f(z) = \frac{1}{z(z-1)} = \frac{1}{z-1} - \frac{1}{z}$$
Now
$$ F(z) =\text{log}(z-1) - \text{log}(z)$$
would be an antiderivative but I don't feel comfortable with the complex logarithm yet. Is my $F(z)$ well-defined on my whole domain? If not, how do I find a well-defined antiderivative?
 A: The expression
$$
F(z)=\log(z−1)−\log(z)
$$
is not well-defined on your domain, because neither $\log(z−1)$
nor $\log(z)$ are holomorphic in $\mathbb{C} \backslash [0,1]$.
One can probably solve that by choosing the arguments of the
logarithms carefully, but it becomes much easier if the right-hand
side is rewritten as $\log \frac{z-1}{z}$.
The Möbius transformation $T(z) = \dfrac{z-1}{z}$ maps $\mathbb{\hat C} \backslash [0,1]$
conformally onto $\mathbb{C} \backslash (-\infty,0]$, so you can
define
$$
F(z) = \log \frac{z-1}{z}
$$
where 
$$\log w = \log|w| + i \arg w \quad (-\pi < \arg w < \pi)
$$ is a holomorphic branch
of the logarithm on $\mathbb{C} \backslash (-\infty,0]$. 
Then $F$ is an "explicit" antiderivate of 
$$
\frac{1}{z-1} - \frac{1}{z}
$$
This can either be verified directly, or you argue as follows:
In any disc $D \subset \mathbb{C} \backslash [0,1]$
$$
  F(z) = \log(z-1) - \log z + C
$$
for some holomorphic branch of $\log(z-1)$ and $ \log z$ in $D$,
and the result follows by differentiation.
A: The solution depends on your closed curve, which poles it does contain in the interior of γ. Since you excluded [0,1] any closed curve will not contain both the poles, or it will contain both and can't have $γ(t)=0$ or $γ(t)=1$. (Curve must have a finite length)
Lets consider:\begin{align}\oint_\gamma f(z)\, dz\end{align} where $f(z)=\frac{1}{z(z-1)}$
Theorem: \begin{align}\oint_\gamma f(z)\, dz =
2\pi i \sum_{k=1}^n \operatorname{I}(\gamma, a_k) 
\operatorname{Res}( f, a_k )\end{align}
In this specific case $n=2$ and $a_1=0,a_2=1$ 
Since a close curve will contain both of them, or none of them, then:
\begin{align}\operatorname{I}(\gamma, a_1)=\operatorname{I}(\gamma, a_2)\end{align} 
Therefore: \begin{align}\oint_\gamma f(z)\, dz =
2\pi i  \operatorname{I}(\gamma, a_1) \sum_{k=1}^2  
\operatorname{Res}( f, a_k )\end{align} 
Theorem 2: Pole order = 1 $\Rightarrow$ \begin{align}Res(f,z_i) = \lim_{z\to z_i} \frac{z-z_i}{f(z)}\end{align}
At $z=0$,\begin{align}Res(f,0) = \lim_{z\to 0} \frac{z}{z(z-1)} = \lim_{z\to 0} \frac{1}{z-1}=-1\end{align} 
At $z=1$ \begin{align}Res(f,1) = \lim_{z\to 1} \frac{z-1}{z(z-1)}=1 \end{align}
Therefore:
\begin{align}\int_\gamma \frac{1}{z(z-1)}dz = 2\pi i  \operatorname{I}(\gamma, a_1) \sum_{k=1}^2  
\operatorname{Res}( f, a_k )=2\pi i  \operatorname{I}(\gamma, a_1) (1-1)=0\end{align}
A: Another possible argument (in response to your above comment):
With the substitution
$$
z = \frac 1w \, ,\quad  dz = -\frac{dw}{w^2}
$$ 
you get
$$
\int_\gamma \frac{1}{z(z-1)} \,dz = \int_{\gamma'} \frac{-1}{\frac 1w(\frac 1w-1)} \,\frac{dw}{w^2} = \int_{\gamma'} \frac{1}{w-1} \, dw
$$
where $\gamma'$ is a closed curve in $D = \Bbb C \setminus [1, \infty) $.
$D$ is simply-connected and $1/(w-1)$ holomorphic in $D$.
It follows from Cauchy's integral theorem that the integral is zero.
