My 'Proof':
If $f\in H(\Omega)$ ,then $f(z)=\sum\limits_{n=0}^{\infty} c_n(z-a)^n,\forall z\in D(a;R)\subset \Omega$ .
Pick $F(z)=\sum\limits_{n=1}^{\infty}\frac{c_{n-1}}{n}(z-a)^n(\forall z\in \Omega)$ , $F(z)$ is well-defined because the expansion of $f$ is unique(not rely on $R$) and converged since $\limsup\limits_{n\to\infty}\frac{1}{^n\sqrt{\frac{c_{n-1}}{n}}}=\limsup\limits_{n\to\infty}\frac{1}{^n\sqrt{c_{n}}}$.
$F(z)\in H(\Omega)$ since analytic function is holomorphic and $F'(z)=f(z)$.
However,this is not true(eg. consider $\int_{\partial \bar D(0;r)}\frac{dz}{z}=2\pi i\neq 0=F(1)-F(0)$).
But if we take $\Omega =\{z|r_1<|z|<r_2,r_1<r<r_2\},f(z)=\frac{1}{z}$ is holomorphic in $\Omega$ and we 'can' find a anti-derivative in $\Omega$ following the steps above.
I'm wondering which step is wrong.Thanks for checking!