If the domain of the function $$f(x) = \log_e\frac{2x+3}{4x^2+x-3} + \cos^{-1}\frac{2x-1}{x+2}$$ is $$(\alpha, \beta]$$, then the value of $$5\beta - 4\alpha$$ is equal to

Domain of $$f(x) = \log_e\frac{2x+3}{4x^2+x-3} + \cos^{-1}\frac{2x-1}{x+2}$$.

For log: $$\frac{2x+3}{4x^2+x-3} > 0$$. Factor denominator: $$4x^2+x-3 = (4x-3)(x+1)$$. Numerator: $$2x+3$$.

$$\frac{2x+3}{(4x-3)(x+1)} > 0$$.

Critical points: $$x = -3/2, -1, 3/4$$. Sign analysis: positive on $$(-3/2, -1) \cup (3/4, \infty)$$.

For $$\cos^{-1}$$: $$-1 \leq \frac{2x-1}{x+2} \leq 1$$, $$x \neq -2$$.

$$\frac{2x-1}{x+2} \leq 1 \Rightarrow \frac{2x-1-x-2}{x+2} \leq 0 \Rightarrow \frac{x-3}{x+2} \leq 0 \Rightarrow x \in (-2, 3]$$.

$$\frac{2x-1}{x+2} \geq -1 \Rightarrow \frac{2x-1+x+2}{x+2} \geq 0 \Rightarrow \frac{3x+1}{x+2} \geq 0 \Rightarrow x \in (-\infty, -2) \cup [-1/3, \infty)$$.

Combined for $$\cos^{-1}$$: $$(-2,3] \cap [(-\infty,-2) \cup [-1/3,\infty)] = [-1/3, 3]$$.

Intersection with log domain: $$[-1/3, 3] \cap [(-3/2,-1) \cup (3/4,\infty)] = (3/4, 3]$$.

So domain = $$(3/4, 3]$$, $$\alpha = 3/4, \beta = 3$$.

$$5\beta - 4\alpha = 15 - 3 = 12$$.

The answer is Option (2): $$\boxed{12}$$.