Does every holomorphic function $f$ defined on an open set $U$ have a primitive? I have read that a holomorphic function in an open disc has a primitive in that disc. Does this mean that for any holomorphic function $f$ in any open set $U,$ that $f$ has a primitive? I say this because I can take a large enough open disc to cover the open set $U$ and find its primitive there. Is this a correct way of thinking? 
 A: No, it's not true.
Consider the function $f(z)=1/z$, which is holomorphic in the open set $\mathbb{C}\setminus\{0\}$, but has no antiderivative.
The key fact is that an open disc is simply connected, but the union of open discs may not be.
The statement is true for simply connected open sets, so it's true that you can find an antiderivative over an open disc around each point, but these may not “glue together”.
On a non simply connected open set there may exist functions not admitting an antiderivative.
Your idea cannot work: consider the situation above: there is no “large enough disc” that covers the given open set. Even if you consider $1/z$ over $D\setminus\{0\}$, where $D$ is the open disc at the origin with radius $1$, you incur in the same problem: whatever point $z_0$ you take in $D\setminus\{0\}$ you have a suitable disc around $z_0$ over which $f(z)=1/z$ has an antiderivative, but this disc can have radius at most $|z_0|$, because it cannot contain $0$.
What you could do is “glueing together” the local antiderivatives, but this doesn't work either. But it's the idea around analytic continuation.
