Let $R' = \mathbb{R}[x]$, $R= \mathbb{R}[x^2 -1] \subset R'$, $P' = (x-1) \subset R'$, and $P = P' \cap R$.
The given hint suggested to consider the element $\frac{1}{x+1} \in R'_{P'}$.
So the following is what I've tried:
Assume that $\frac{1}{x+1}$ were integral over $R_P$, i.e., there exist $a_{n-1}, ..., a_0 \in R_P$ such that
$(\frac{1}{x+1})^n + a_{n-1} (\frac{1}{x+1})^{n-1}+ ... + a_0 = 0$
Then multiplying both sides with $(x^2 - 1)^n$ gives us
$(x-1)^n + a_{n-1} (x^2-1) (x-1)^{n-1} + ... + a_0 (x^2 - 1)^n=0$
Then plugging in $x=-1$, it follows that $(-2)^n = 0 $,
which is a contradiction.
Here is the point:
I'm not sure if it is okay to plug in $x=-1$, since $R'_{P'}$ is well-defined near $x=1$, not $x=-1$.
Edit: This approach seems to be wrong, since the $x+1$ is at the denominator , plugging in $x=-1$ is not allowed.
Could you give me a hint? Thank you.
$b=(x-1)^n + a_{n-1} (x^2-1) (x-1)^{n-1} + ... + a_0 (x^2 - 1)^n$ is an element of $\Bbb{R}[x^2-1]_{(x^2-1)}[x-1]$ which clearly has a well-defined $\Bbb{R}$-valued homomorphism $x\to -1$. It shows that $b\ne 0$ in $\Bbb{R}[x^2-1]_{(x^2-1)}[x-1]$ thus in $\Bbb{R}(x)$ which is a contradiction.