proof of the Cauchy integral formula 
$ D=D_{r}(c), r > 0 .$ Show that if $f$ is continuous in $\overline{D}$ and holomorphic in $D$, then for all $z\in D$: $$f(z)=\frac{1}{2\pi i} \int_{\partial D}\frac{f(\zeta)}{\zeta - z} d\zeta$$ 

I don't understand this question because I don't see how it is different to the special case of the Cauchy Integral formula. I would be very glad if somebody could tell me what the difference is and how to show that it is true. 

If I understand correctly, $D_r(c)$ is the open ball centered in $c$ with radius $r$? If this is the case, the difference between the two is that above your $c$ is fixed, and in the special case your $c$ "moves" with the ball.
Fix $c$ then; we want to show that for every $z\in D_r(c)$ we have that
$$f(z)=\frac{1}{2\pi i}\int_{\partial D_r(c)}\frac{f(\zeta)}{\zeta-z}d\zeta$$
(we assume that $f$ is holomorphic in the ball).
Well, if we choose $s>0$ such that $D_s(z)\subseteq D_r(c)$, then by the special case we have that
$$f(z)=\frac{1}{2\pi i}\int_{\partial D_s(z)}\frac{f(\zeta)}{\zeta-z}d\zeta,$$
and we see that $\partial D_s(z)$ is homologous to $D_r(c)$. We then have that
$$f(z)=\frac{1}{2\pi i}\int_{\partial D_s(z)}\frac{f(\zeta)}{\zeta-z}d\zeta=\frac{1}{2\pi i}\int_{\partial D_r(c)}\frac{f(\zeta)}{\zeta-z}d\zeta.$$
Does that answer your question? Is that what your question was?
The nice thing about the "general" Cauchy formula is that the curve that we're integrating over no longer depends on the point you want to evaluate.
