Integral Closure of Ring $k[t,s]/(s^2 -t^2(1+t))$ 
Let $X$ be the affine curve $\operatorname{Spec}k[t,s]/(s^2 -t^2(1+t))$.
How to verify that the normalization $X'$ of $X$ is $\operatorname{Spec}k[u]$ with $u:= s/t$ and $u^2=1+t$?

We know that since the normalization $n: X' \to X$ is an affine map, $X'$ is an affine scheme $\operatorname{Spec}A$ with $A$ integral closure of $k[t,s]/(s^2 -t^2(1+t))$ in $\operatorname{Frac}\bigl(k[t,s]/(s^2 -t^2(1+t))\bigr)$.
Is there any standard strategy to calculate integral closure of rings $R:= k[x,y]/(P(x,y))$ in $\operatorname{Frac}(R)$?

Prove that the normalisation of $A=k[X,Y]/(Y^2-X^2-X^3)$ is $k[t]$ where $t=Y/X$ (Reid, Exercise 4.5)
Characterizing the field of fractions of $\mathbb Q[x,y]/(x^2+y^2-1)$.
Working out the normalization of $\mathbb C[X,Y]/(X^2-Y^3)$
$\mathbb{Q}[x,y]/(x^2+y^2)\cong \mathbb{Q}[y,yi]$?
As in the above links, I think it seems easier when we find an isomorphic subring in a polynomial ring. In your case,
we first define a ring homomorphism, $\phi : k[t,s]\rightarrow k[u]$ given by $\phi(t)=u^2-1,\phi(s)=u(u^2-1)$. Then $\ker\phi=(s^2-t^2(1+t))$ and $k[t,s]/(s^2-t^2(1+t))\cong k[u^2-1,u(u^2-1)]\hookrightarrow k[u]$ and $\mathrm{Frac}(k[u^2-1,u(u^2-1)]=\mathrm{Frac}(k[u])=k(u)$. So we first  may need to parametrize variables very well. Proving the equality of the kernel is not really obvious to me and the method used in the following link and comments on that would be helpful 
The kernel of $\mathbb{Q}[x,y]\rightarrow \mathbb{Q}(t)$ is $(x^2+y^2-1)$.
Another strategy is
The field of fractions of $\mathbb{R}[x,y]/(x^2+y^2-1)$ is $\mathbb{R}(x)[y]/(x^2+y^2-1)$
Showing $F[x,y]/(ax^2+by^2-1)$ is a Dedekind domain
