Quadratic extension of $\mathbb{Q}(X)$ Consider the ring $\mathbb{Q}[X]$ of polynomials in $X$ with coefficients in the field of rational numbers. Consider the quotient field $\mathbb{Q}(X)$ and let $K$ be a quadratic extension of $\mathbb{Q}(X)$ given by $K:=\mathbb{Q}(X)[d]$.
Let $O_{K}$ be the integral closure of $\mathbb Q[X]$ in $K$. Certainly $O_{K}$ contains both $\mathbb{Q}[X]$ and $d$, hence
$$ O_{K}\supseteq \mathbb{Q}[X][d].$$
Do you think the previous inclusion is strict, or the equality holds?
EDIT. The conclusion $O_{K}\supseteq \mathbb{Q}[X][d]$ is false. So the question is: who is $O_{K}$?
 A: You write:    "Certainly $O_{K}$ contains both $\mathbb{Q}[X]$ and $d$, hence
$ O_{K}\supseteq \mathbb{Q}[X][d]$"  
But this is not true:
Take $d=\sqrt \frac {1}{X}$, a root ot $T^2-\frac {1}{X}\in \mathbb Q(X)[T]$.
Then $d$ is not integral over $\mathbb Q[X]$ because if it were, then also $d^2=\frac {1}{X}$, would be integral over $\mathbb Q[X]$ and we both know very well that this is absurd.
So, $d\notin O_K$ and thus   $ O_{K}\supseteq \mathbb{Q}[X][d]$ does not hold.       
Edit
As for the computation of $O_K$, here is a result which may be of help: it is given as Theorem 9.2 (page 65) in Matsumura's Commutative ring theory.  

Theorem
   Let $A$ be an integrally closed domain and $K$ an algebraic extension of its field of fraction  $F=Frac(A)$.
   Then an element $k\in K$ is integral over $A$ iff its minimal polynomial over $F$, the monic polynomial $f(T)=Irr(k,F,T)\in F[T]$,   has its coefficients in $A$.  

In your case, if  you suppose for example that $d=\sqrt {P(X)}$ with $P(X)\in \mathbb Q|X]$ a square-free polynomial, the theorem above will allow you to prove that $O_K$ is the ring $\mathbb Q[X][d]$ (details in Matsumura).
(The above class of examples generalizes the result of your preceding question but doesn't completely solve the  problem of calculating $O_K$ for general quadratic extensions $K$  of $\mathbb Q(X)$).
