Let y=y(x) be the solution curve of the differential  equation

$\frac{dy}{dx}+\frac{1}{x^2-1}y={\left(\frac{x-1}{x+1}\right)}^{\frac{1}{2}},x>1$passing through the point $\left(2,\sqrt{\frac{1}{3}}\right)$ Then $\sqrt{7} y\left(8\right)$ is equal to

$$\begin{gathered} \frac{dy}{dx}+\frac{1}{x^2-1}y={\left(\frac{x-1}{x+1}\right)}^{\frac{1}{2}}, \mathrm{is} \mathrm{a} \mathrm{linear} \mathrm{differential} \mathrm{equation} \mathrm{of} \mathrm{the} \mathrm{form} \\ \frac{dy}{dx}+py =Q \\ I.F.=e^{\int pdx}=e^{\int \frac{1}{x^2-1}dx}=e^{\frac{1}{2}\log \frac{x-1}{x+1}}={\left(\frac{x-1}{x+1}\right)}^{\frac{1}{2}} \\ \mathrm{solution} \mathrm{is} \mathrm{given} \mathrm{by} \\ \mathrm{y}{\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)}^{\frac{1}{2}}=\int {\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)}^{1}dx \\ \Rightarrow \mathrm{y}{\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)}^{\frac{1}{2}}=\int \left(1-\frac{2}{\mathrm{x}+1}\right)\mathrm{dx} \\ \Rightarrow \mathrm{y}{\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)}^{\frac{1}{2}}=x-2{\log }_e\left|x+1\right|+C \\ Cuvve passes through\left(2,\frac{1}{\sqrt{3}}\right) \\ \Rightarrow C=2{\log }_{e}3-\frac{5}{3} \\ at x=8, \\ y\left(8\right){\left(\frac{8-1}{8+1}\right)}^{\frac{1}{2}}=8-2\log 9+2\log 3-\frac{5}{3} \\ \Rightarrow \sqrt{7}y\left(8\right)=3\left(\frac{19}{3}-2{\log }_{e}3\right) \\ \sqrt{7}y\left(8\right)=19-6 {\log }_{e}3 \end{gathered}$$