If $y=y\left(x\right)$ is the solution of the differential equation $\frac{dy}{dx}+\frac{4x}{x^2-1}y=\frac{x+2}{{\left(x^2-1\right)}^{5/2}}$ $x>1$ such that  $y\left(2\right)=\frac{2}{9}\log \left(2+\sqrt{3}\right)$ and $y\left(\sqrt{2}\right)=\alpha {\log }_e\left(\sqrt{\alpha }+\beta \right)+\beta -\sqrt{r}$ $\alpha ,\beta ,r\epsilon N$ then  $\alpha \beta r$ is equal to ..... 

I.F $=e^{2\int \frac{2x}{x^2-1}dx}$

$=e^{2\log x^2-1}={\left(x^2-1\right)}^2$

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

$=\int \frac{x+2}{\sqrt{x^2-1}}dx=\frac{1}{2}$

$\int \frac{x+2}{x-1}dx+2\int \frac{dy}{\sqrt{x^2-1}}$

$y{\left(x^2-1\right)}^2=\sqrt{x^2-1}+2\log$

$\left|x+\sqrt{x^2-1}\right|+c$

put  $x=2y=\frac{2}{9}\log \left(2+\sqrt{3}\right)$

$\frac{2}{9}\log \left(2+\sqrt{3}\right){\left(4-1\right)}^2=$

$\sqrt{4-1}+2\log \left|2+\sqrt{3}\right|+c$

$c=-\sqrt{3}$

$y{\left(x^2-1\right)}^2=\sqrt{x^2-1}+2\log$

$\left|x+\sqrt{x^2-1}\right|-\sqrt{3}$

put $x=\sqrt{2}\Rightarrow y=1+2$

$\log \left(\sqrt{2}+1\right)-\sqrt{3}$

$y=2\log \left(\sqrt{2}+1\right)+1-\sqrt{3}$

Here $\alpha =2\beta =1\gamma =3$

$\alpha \beta \gamma =6$