The following proof is based on the assumption that $A$ is Noetherian and is integrally closed (in its field of fractions $K$):
(1) If $L$ is a finite separable extension of $K$, then the bilinear form $(x,y)\to \text{Tr}_{L/K}(xy)$ on $L$ is nondegenerate. (Exercise)
(2) Choose a basis of $L$ over $K$. We can assume, after possibly scaling by elements of $K$, that this basis consists entirely of elements of $B$ (the integral closure of $A$ in $L$). (Exercise) Let $(v_1,\dots,v_n)$ be this basis of $L$ (as a vector space over $K$).
(3) We know by (1) that there is a basis $(w_1,\dots,w_n)$ of $L$ over $K$ "dual" to the basis $(v_1,\dots,v_n)$ of $L$ over $K$. More precisely, there is a basis $(w_1,\dots,w_n)$ of $L$ over $K$ such that $\text{Tr}_{L/K}(v_iw_j)=\delta_{ij}$ for all $1\leq i,j\leq n$ (where $\delta_{ij}$ denotes the Kronecker delta). (Exercise)
(4) Let $x\in B$ and write $x=\Sigma_{j=1}^n x_jw_j$ where $x_j\in K$ for all $1\leq j\leq n$. Note that $xv_i\in B$ for each $1\leq i\leq n$ and thus $\text{Tr}_{L/K}(xv_i)\in A$. (Exercise) However, $\text{Tr}_{L/K}(xv_i)=\Sigma_{j=1}^n x_j\text{Tr}_{L/K}(w_jv_i)= x_i$. Therefore, $x_i\in A$ for all $1\leq i\leq n$ and $x\in \Sigma_{i=1}^n Av_i$.
(5) Since $A$ is Noetherian, it follows that $B$ is a finitely generated $A$-module. (Exercise)
An interesting corollary that is fundamental in algebraic number theory:
Corollary Let $A$ be a Dedekind domain and let $K$ be its field of fractions. If $L$ is a finite separable extension of $K$, then the integral closure of $A$ in $L$ is also a Dedekind domain.
Proof. We know that $B$ is Noetherian since it is a finitely generated $A$-module (and $A$ is Noetherian). Also, $B$ is integrally closed in $L$. (Exercise) Therefore, all that remains to show is that every non-zero prime ideal of $B$ is maximal. (Exercise) Q.E.D.
I hope this helps!
The fraction field $K$ has the structure of an $R$-module, and for any maximal ideal $\mathfrak{m}$ of $R$, it also naturally has the structure of an $R_\mathfrak{m}$-module. If $K$ is finitely generated as an $R$-module, then $K$ must also be finitely generated as an $R_\mathfrak{m}$-module because $R_\mathfrak{m}\supseteq R$. Thus we are reduced to the case when $R$ is local.
Let $R$ be a local domain with maximal ideal $\mathfrak{m} \neq 0$. Clearly, $\mathfrak{m}K=K$. But the Jacobson radical of $R$ is just $\mathfrak{m}$, so if $K$ is finitely generated as an $R$-module, then Nakayama's lemma implies that $K=0$, which is a contradiction.
Best Answer
Eisenbud has a discussion of this question in Section 13.3 of Commutative Algeba with a View Toward Algebraic Geometry, as does Matsumura in Section 33 of Commutative Ring Theory. If $R$ is a domain that is a finitely generated algebra over a field, then the integral closure of $R$ is finitely generated as an $R$-module. If $R$ is a complete Noetherian local domain, then the integral closure of $R$ is finitely generated as an $R$-module.
In 1935/6, Akizuki and Schmidt (separately) constructed examples of one-dimensional local Noetherian domains $R$ whose integral closure is not finitely generated as an $R$-module. There are many more examples known today. For instance, Goodearl and Lenagan gave a nice construction in 1989 of such examples here, though they are still a bit involved.
The Krull-Akizuki Theorem says that in the examples $R$ above (of dimension $1$), the integral closure is a Noetherian ring; this is also true if $R$ is $2$-dimensional. In 1953, at the link given by @mathmath, Nagata constructed a $3$-dimensional local Noetherian domain whose integral closure is not Noetherian.