If $$y = y(x)$$ is the solution of the differential equation $$\frac{dy}{dx} + \frac{4x}{x^2-1}y = \frac{x+2}{(x^2-1)^{5/2}}$$, $$x \gt 1$$ such that $$y(2) = \frac{2}{9}\log_e 2 + \sqrt{3}$$ and $$y\sqrt{2} = \alpha\log_e(\sqrt{\alpha} + \beta) + \beta - \sqrt{\gamma}$$, $$\alpha, \beta, \gamma \in \mathbb{N}$$, then $$\alpha\beta\gamma$$ is equal to _____.

$$\frac{dy}{dx} + \left( \frac{4x}{x^2 - 1} \right) y = \frac{x+2}{(x^2 - 1)^{5/2}}$$

$$\text{I.F.} = e^{\int \frac{4x}{x^2 - 1} \, dx} = e^{2 \ln\vert{}x^2 - 1\vert{}} = (x^2 - 1)^2$$

$$y \cdot (x^2 - 1)^2 = \int \frac{x+2}{(x^2 - 1)^{5/2}} \cdot (x^2 - 1)^2 \, dx$$

$$y \cdot (x^2 - 1)^2 = \int \frac{x+2}{\sqrt{x^2 - 1}} \, dx = \int \frac{x}{\sqrt{x^2 - 1}} \, dx + 2\int \frac{1}{\sqrt{x^2 - 1}} \, dx$$

$$y \cdot (x^2 - 1)^2 = \sqrt{x^2 - 1} + 2\ln\vert{}x + \sqrt{x^2 - 1}\vert{} + C$$

$$y(2) \cdot (4 - 1)^2 = \sqrt{3} + 2\ln\vert{}2 + \sqrt{3}\vert{} + C$$

$$9 \left( \frac{2}{9}\log_e 2 + \frac{\sqrt{3}}{9} \right) = \sqrt{3} + 2\ln(2+\sqrt{3}) + C$$

$$2\log_e 2 + \sqrt{3} = \sqrt{3} + 2\ln(2+\sqrt{3}) + C \implies C = \log_e 4 - \ln(7 + 4\sqrt{3})$$

$$y \cdot (x^2 - 1)^2 = \sqrt{x^2 - 1} + 2\ln\left( \frac{x + \sqrt{x^2 - 1}}{2+\sqrt{3}} \right) + 2\log_e 2$$

$$y(\sqrt{2}) \cdot (2 - 1)^2 = \sqrt{1} + 2\ln\left( \frac{\sqrt{2} + 1}{2+\sqrt{3}} \right) + 2\log_e 2$$

$$y(\sqrt{2}) = 2\log_e(\sqrt{2}+1) + 1 - \sqrt{3}$$

$$\alpha\beta\gamma = 2 \cdot 3 \cdot 1 = 6$$