Is an integer extension of a ring the integral closure of this ring in some extension? This is a question that I just came up with, so it may be completely stupid and I sincerely apologize if that's the case.
The question is; Take $A$ to be an integral domain. Let $B$ be an integral extension of $A$ (i.e. every $b\in B$ satisfies an equation of the form $b^n+a_{n-1}b^{n-1}+\dots+a_0=0$ with $a_0,\dots,a_{n-1}\in A$). Is it true that there is an extension of the field of fractions of $A$, say $L$, such that $B$ equals the integral closure of $A$ in $L$ (i.e. the elements of $L$ which satisfy a polynomial equation as above).
I am very sure that the answer is no but I wonder if there are some specific cases where it may hold or any nice counterexample.

$\newcommand{\Q}{\Bbb Q}
\newcommand{\N}{\Bbb N}
\newcommand{\R}{\Bbb R}
\newcommand{\Z}{\Bbb Z}
\newcommand{\C}{\Bbb C}
\newcommand{\A}{\Bbb A}
\newcommand{\Frac}{\mathrm{Frac}}
$
No. Take the integral extension 
$$A= \Z \subset B= \Z[\sqrt 5].$$ 
If there was an extension $L \supset \Frac(A) = \Q$ such that $B$ equals the integral closure $O$ of $A$ in $L$, then $L$ would be finite and $B=O$ would be a Dedekind domain, since $A$ is (see Prop. I.8.1 in Neukirch's Algebraic number theory). But $B$ is not integrally closed, so it can't be Dedekind.
