What is $d(\mathbb{Q}(\sqrt{2},\sqrt{3}))$? $K=\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$ and $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2}+\sqrt{3}):\mathbb{Q}]=4$. We also know the conjugates of $\sqrt{2}+\sqrt{3}$ are,
$$x_1=\sqrt{2}+\sqrt{3}$$
$$x_2=\sqrt{2}-\sqrt{3}$$
$$x_3=-\sqrt{2}+\sqrt{3}$$
$$x_4=-\sqrt{2}-\sqrt{3}$$
$\mathbb{Q}(\sqrt{2}+\sqrt{3})=\{a+b(\sqrt{2}+\sqrt{3})+c(\sqrt{2}+\sqrt{3})^2+d(\sqrt{2}+\sqrt{3})^3|a,b,c,d \in \mathbb{Q}\}$. Thus an integral basis of $K$ is,
$$S=\{1,\sqrt{2}+\sqrt{3},(\sqrt{2}+\sqrt{3})^2,(\sqrt{2}+\sqrt{3})^3\}$$
$$S=\{1,\sqrt{2}+\sqrt{3},5+2\sqrt{6},(\sqrt{2}+\sqrt{3})^3\}$$
and hence,
$$d(K)=d(\mathbb{Q}(\sqrt{2}+\sqrt{3}))=D(S)$$
$$D(S)=\begin{vmatrix}
1 & \sqrt{2}+\sqrt{3} & (\sqrt{2}+\sqrt{3})^2 & (\sqrt{2}+\sqrt{3})^3\\
1 & \sqrt{2}-\sqrt{3} & (\sqrt{2}-\sqrt{3})^2 & (\sqrt{2}-\sqrt{3})^3\\
1 & -\sqrt{2}+\sqrt{3} & (-\sqrt{2}+\sqrt{3})^2 & (-\sqrt{2}+\sqrt{3})^3\\
1 & -\sqrt{2}-\sqrt{3} & (-\sqrt{2}-\sqrt{3})^2 & (-\sqrt{2}-\sqrt{3})^3
\end{vmatrix}^2$$
$$D(S)=\begin{vmatrix}
1 & \sqrt{2}+\sqrt{3} & 5+2\sqrt{6} & 11\sqrt{2}+9\sqrt{3}\\
1 & \sqrt{2}-\sqrt{3} & 5-2\sqrt{6} & 11\sqrt{2}-9\sqrt{3}\\
1 & -\sqrt{2}+\sqrt{3} & 5-2\sqrt{6} & -11\sqrt{2}+9\sqrt{3}\\
1 & -\sqrt{2}-\sqrt{3} & 5+2\sqrt{6} & -11\sqrt{2}-9\sqrt{3}
\end{vmatrix}^2$$
Using Mathematica,
$$D(S)=2677248 + 1078272\sqrt{6}$$
Why am I not getting an integer answer?
Appararently I don't know how to type, $D(S)= 147456$

There are two issues here: First, you seem to have made a data entry error, which resulted in a non-integer answer. Second, $\mathbb{Z}[\sqrt{2} + \sqrt{3}]$ isn't integrally closed, so you're computing the discriminant of some non-maximal order of $\mathbb{Q}(\sqrt{2} + \sqrt{3})$.
The following SageMath code verifies that $\mathbb{Z}[\sqrt{2} + \sqrt{3}]$ in fact has index $8$ in the maximal order:
sage: K.<b> = NumberField((sqrt(2) + sqrt(3)).minpoly())
sage: R = K.order(b)
sage: OK = K.maximal_order()
sage: R.index_in(OK)
8

Thus, the discriminant of this order is $8^2$ times the discriminant of the number field. (For reference, this is number field 4.4.2304.1 in the LMFDB.)
