We can solve the given differential equation using an integrating factor.

The integrating factor is given by :

$$ \mu(x) = e^{\int \frac{4x}{x^2 - 1} dx} = e^{2\ln(x^2 - 1)} = (x^2 - 1)^2 $$

Multiplying both sides of the differential equation by $\mu(x)$, we get :

$$(x^2-1)^2 \frac{d y}{d x} + 4x y = \frac{x+2}{\left(x^{2}-1\right)^{\frac{1}{2}}}$$

We can rewrite the left-hand side using the product rule:

$$\frac{d}{dx} \left((x^2-1)^2 y\right) = \frac{x+2}{\left(x^{2}-1\right)^{\frac{1}{2}}}$$

Integrating both sides with respect to $x$, we get:

$$(x^2-1)^2 y = \int \frac{x+2}{\left(x^{2}-1\right)^{\frac{1}{2}}} dx = \sqrt{x^2-1}+2 \ln\left|x+\sqrt{x^2-1}\right|+C$$

where $C$ is the constant of integration. Using the initial condition $y(2) = \frac{2}{9} \log_e(2+\sqrt{3})$, we can solve for $C$: