$u^{-1}$ is integral over $R\subset S$ if and only if $u^{-1}\in R[u]$ I need to prove the following: 
Let $R\subset S$ be a commutative ring and $u$ be any invertible element in $S$. Then $u^{-1}$ is integral over $R$ is and only if $u^{-1}\in R[u].$
Proof: Let $u^{-1}$ is integral over $R \implies \exists\  $ a monic polynomial $f(x)\in R[x]$ such that $f(u^{-1})=0$. If 
\begin{align*}
f(x) &=x^n+r_{n-1}x^{n-1}+\ldots+r_1x+r_0\\
\implies f(u^{-1}) &=u^{-n}+r_{n-1}u^{-n+1}+\ldots+r_1u^{-1}+r_0=0\\
\implies &u^{-1}+r_{n-1}+r_{n-2}u+\ldots+r_1u^{n-2}+r_0u^{n-1}=0
\end{align*}
Therefore, $u^{-1}\in R[u]$. 
Now how to do the converse.

Suppose $u^{-1} \in R[u]$. Then there exists some polynomial $f(x) \in R[x]$ such that $u^{-1}=f(u)$, or equivalently
$$uf(u)-1=0$$
Write $f(x)= \sum_{i=0}^n a_ix^i$. Then
$$0=\sum_{i=0}^n a_iu^{i+1}-1 = u^{n+1} \cdot (u^{-(n+1)}+a_0u^{-n} + a_1 u^{-(n-1)}+ \dots + a_n)$$
Since $u^{n+1}$ is a unit, you have
$$u^{-(n+1)}+a_0u^{-n} + a_1 u^{-(n-1)}+ \dots + a_n=0$$
which is an integral relation of $u^{-1}$ over $R$.
