The given inequality is ((x+1)²+3)/((x+3)(x−2)) < 0.

The given inequality is $\frac{(x+1)^2+3}{(x+3)(x-2)} < 0$.
Consider the numerator: $(x+1)^2+3$.
Since $(x+1)^2$ is always greater than or equal to 0 for any real number $x$, adding 3 to it means that $(x+1)^2+3$ is always greater than or equal to 3.
Therefore, the numerator $(x+1)^2+3$ is always positive.

For the entire fraction to be less than 0 (negative), and since the numerator is always positive, the denominator must be negative.
So, we must have $(x+3)(x-2) < 0$.

To find when $(x+3)(x-2) < 0$, we find the roots of the quadratic expression $(x+3)(x-2) = 0$. The roots are $x = -3$ and $x = 2$.
This is a parabola opening upwards. The expression $(x+3)(x-2)$ will be negative when $x$ is between its roots.
Thus, the solution to $(x+3)(x-2) < 0$ is $-3 < x < 2$.

The set of values for $x$ is $\{x : -3 < x < 2\}$.
This matches option D.

The final answer is $\text{D}$.