Familiarity breeds contempt!

$\large \displaystyle \frac{dy}{dx} = \frac{1}{x\sqrt{x^2-1}}$

$\large \displaystyle y = \int \frac{1}{x\sqrt{x^2-1}} dx$

$\large \displaystyle \color{#3D99F6} u = \sqrt{x^2-1} \rightarrow x = \pm \sqrt{u^2+1}$

$\large \displaystyle \color{#3D99F6} du = \frac{x}{\sqrt{x^2-1}} dx \rightarrow du = \pm \frac{\sqrt{u^2+1}}{u} dx$

$\large \displaystyle y = \int \frac{1}{u\sqrt{u^2+1}} \cdot \frac{u}{\sqrt{u^2+1}} du$

$\large \displaystyle y = \int \frac{1}{u^2+1}du$

$\large \displaystyle y = \arctan(u) + C$

$\large \displaystyle y = \arctan(\sqrt{x^2-1}) + C$

Since it passes through the point $(-2, -2)$ :

$\large \displaystyle -2 = \arctan(\sqrt{3}) + C$

$\large \displaystyle -2 = \frac{\pi}{3} + C$

$\large \displaystyle \color{#3D99F6} C = -2 - \frac{\pi}{3}$

So:

$\color{#3D99F6} \boxed{ \large \displaystyle y = \arctan(\sqrt{x^2-1}) -2 - \frac{\pi}{3} }$

Which passes through points $(A)$ and $(C)$ only.