I would like to know whether my argument is correct or not. We consider
\begin{align} f: D \to \mathbb{C}, z \to \frac{1}{1+z^3} \quad \text{and} \quad \int_{\gamma} f(z) \mathrm{d}z \end{align} where $\gamma : [-\pi; \pi] \to \mathbb{C}, \ t \to 2e^{it}$ and $D = \{ z \in \mathbb{C} : |z| > 1 \}$. I would like to decide whether or not $f$ does have a holomorphic antiderivative in $D$.
I argued that the zeros of $1+z^3$ are given by $\{-1, \frac{1}{2} \pm i \frac{\sqrt{3}}{2}\}$ and that they are all not contained in $D$. Therefore $f$ is holomorphic in $D$. Since $\gamma$ is a closed curve and completely in $D$ we have that
\begin{align} \int_{\gamma} f(z) \ \mathrm{d}z = 0. \end{align}
Since $D \subseteq \mathbb{C}$ is open and $f$ is continuous $f$ does have a holomorphic antiderivative in $D$.