Is $\mathbb{Z}[[t]][x]/(x^2-(1+t))$ integrally closed? Since $\mathbb{Z}$ is a PID, $\mathbb{Z}[[t]]$ is a UFD. Note that the square root of $1+t$ is:
$$1 + \frac{t}{2} -\frac{t^2}{8}+\cdots$$
Thus, the polynomial $x^2-(1+t)$ has no roots in $\mathbb{Z}[[t]]$, so it is irreducible, hence prime, and thus $\mathbb{Z}[[t]][x]/(x^2-(1+t))$ is an integral domain.
Is it integrally closed? What is its ramification locus over $\mathbb{Z}[[t]]$?
EDIT: The ramification locus certainly contains $(2)$ and $(1+t)$. Could it be ramified elsewhere?
 A: Here is my argument that it is integrally closed. I don't understand the second part of your question (sorry), is it a Dedekind domain ?
An integral element must be of the form
$$\frac{a+b\sqrt{1+t}}{2}$$ with $a,b \in \mathbb{Z}[[t]]$. This follows from the standard algebraic number theory argument, see also Example 4 on page 65 of Matsumura, Commutative Ring Theory. Further this will be integral iff $$\frac{a^2-b^2(1+t)}{4}\in \mathbb{Z}[[t]].$$
So we may assume that all the coefficients of the power series $a$ and $b$ are $0, \pm 1$. Further we may remove any common factors of $t$ in $a$ and $b$ to assume without loss that at least one of the constant terms is non-zero.
Now the constant term of $\frac{a^2-b^2(1+t)}{4}$ is $$\frac{a_0^2-b_0^2}{4}$$ and if this is an integer then $a_0^2=b_0^2$, in particular both are non-zero. So $b_0\neq 0$.
Next the coefficient of $t$ is $$\frac{2a_0a_1-2b_0b_1-b_0^2}{4}.$$  
Since $b_0=\pm 1$ this is never an integer. Thus we see that this expression cannot be integral. 
