Atiyah-Macdonald, Exercise 5.4 I was having some trouble with the following exercise from Atiyah-Macdonald.

Let $A$ be a subring of $B$ such that $B$ is integral over $A$. Let $\mathfrak{n}$ be a maximal ideal of $B$ and let $\mathfrak{m}=\mathfrak{n} \cap A$ be the corresponding maximal ideal of $A$. Is $B_{\mathfrak{n}}$ integral over $A_{\mathfrak{m}}$?

The book gives a hint which serves as a counter-example. Consider the subring $k[x^{2}-1]$ of $k[x]$ where $k$ is a field, and let $\mathfrak{n}=(x-1)$. I am trying to show that $1/(x+1)$ could not be integral over $k[x^{2}-1]_{\mathfrak{n}^{c}}$.
I have understood why this situation serves as a counterexample. But I am essentially stuck at trying to draw a contradiction. A hint or any help would be great. 

Maybe you already noticed that $\mathfrak n^c=(x^2-1)$. Now apply the definition of integrality and after clearing the denominators you get $\sum_{i=0}^n a_is_i(x+1)^{n-i}=0$ with $a_i\in A$, $a_n=1$, and $s_i\in A-\mathfrak n^c$. Then $x+1\mid s_n$ (in $B$), so $s_n\in (x+1)B\cap A=(x^2-1)$, a contradiction.
