The solution of differential equation \(\sqrt {x + 1 }\) - \(\sqrt {x - 1 }\)\(\frac{dy}{dx}\) = 0 is

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

\(\sqrt {x + 1 }\) - \(\sqrt {x - 1 }\)\(\frac{dy}{dx}\) = 0

$⇒ \int\frac{\sqrt{x+1}}{\sqrt{x-1}}dx=\int dy$

$⇒ \int\frac{x+1}{\sqrt{x^2-1}}dx=y$

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

$y=\sqrt{x^2-1}+log|x+\sqrt{x^2-1}|+C$

Option 1 is correct.