To find the domain of the function $F(x)$, we need to determine the set of all real numbers $x$ for which the function is defined. The function is composed of two main parts, so we must find the domain for each part and then take their intersection.
Let $F(x) = f(x) + g(x)$, where:
Domain of $f(x) = \sqrt{\ln(x-1)}:$
For this part to be defined, two conditions must be met:
The argument of the logarithm must be positive:
$$x - 1 > 0 \implies x > 1$$
The expression inside the square root must be non-negative:
$$\ln(x-1) \ge 0$$
Since $\ln(1) = 0$, we have:
$$x - 1 \ge 1 \implies x \ge 2$$
Taking the intersection of $x > 1$ and $x \ge 2$, the domain for the first part is:
$$D_1 = [2, \infty)$$
Domain of $g(x) = \ln\left(\dfrac{\sqrt{x+1}-2}{x^2-5x+6}\right):$
For this part to be defined, three conditions must be met:
The inner square root must be real:
$$x + 1 \ge 0 \implies x \ge -1$$
The denominator cannot be zero:
$$x^2 - 5x + 6 \neq 0$$
$$(x-2)(x-3) \neq 0 \implies x \neq 2 \text{ and } x \neq 3$$
The argument of the logarithm must be strictly positive:
$$\frac{\sqrt{x+1}-2}{(x-2)(x-3)} > 0$$
Let's use a sign chart to analyze the ratio. We need to find the signs of the numerator and the denominator on the intervals defined by their roots, keeping in mind $x \ge -1$.
Numerator: Let $N(x) = \sqrt{x+1} - 2$.
$N(x) = 0$ when $\sqrt{x+1} = 2 \implies x+1 = 4 \implies x = 3$.
For $x < 3$, $N(x)$ is negative.
For $x > 3$, $N(x)$ is positive.
Denominator: Let $D(x) = (x-2)(x-3)$.
The roots are $x=2$ and $x=3$. It is a parabola opening upwards.
For $x < 2$ or $x > 3$, $D(x)$ is positive.
For $2 < x < 3$, $D(x)$ is negative.
Now, let's look at the sign of the fraction $\frac{N(x)}{D(x)}$ across the valid intervals (where $x \ge -1$):
Interval $[-1, 2)$: Numerator is $(-)$, Denominator is $(+)$. Fraction is $(-)$. (Invalid)
Interval $(2, 3)$: Numerator is $(-)$, Denominator is $(-)$. Fraction is $(+)$. (Valid)
Interval $(3, \infty)$: Numerator is $(+)$, Denominator is $(+)$. Fraction is $(+)$. (Valid)
Thus, the domain for the second part is:
$$D_2 = (2, 3) \cup (3, \infty)$$
Final Intersection :
The domain of the entire function $F(x)$ is the intersection of $D_1$ and $D_2$:
$$D = D_1 \cap D_2$$
$$D = [2, \infty) \cap \left( (2, 3) \cup (3, \infty) \right)$$
Because $x=2$ is excluded from $D_2$ (it makes the denominator zero), it cannot be in the final domain.
$$\boxed{D = (2, 3) \cup (3, \infty)}$$
