jmc

Triangles $△ABC$ and $△ADC$ are both right and share hypotenuse $AC$, which has length $3$. Thus we have $$A{B}^2+B{C}^2=A{D}^2+D{C}^2=3^2=9.$$The only possible integer values for $AB,$ $BC,$ $AD,$ or $DC$ are $1$ and $2$. Thus we may assume that one leg of $△ABC$ has length $1$ and one leg of $△ADC$ has length $2$ (it doesn't matter if the labels $B$ and $D$ have to be swapped to make this true).

If one leg of $△ABC$ has length $1,$ then the other leg has length $\sqrt{3^2-1^2}=\sqrt{8}=2\sqrt{2}$. If one leg of $△ADC$ has length $2,$ then the other leg has length $\sqrt{3^2-2^2}=\sqrt{5}$. Thus, quadrilateral $ABCD$ is divided by its diagonal $AC$ into right triangles of area $\frac{1\cdot 2\sqrt{2}}{2}=\sqrt{2}$ and $\frac{2\cdot \sqrt{5}}{2}=\sqrt{5}$. So, the area of quadrilateral $ABCD$ is $\boxed{\sqrt{2}+\sqrt{5}}$.