= Solution
On an interval avoiding $x=\pm1$, <separation of variables> gives, for $y\ne-1$,
$$
\ln|y+1|=\int\frac{dx}{x^2-1}=\frac12\ln\left|\frac{x-1}{x+1}\right|+C_0.
$$
Absorb the sign into the integration constant and include the <equilibrium solution> $y=-1$ by allowing $C=0$. The complete real family is
$$
\boxed{y=-1+C\sqrt{\left|\frac{x-1}{x+1}\right|},\qquad C\in\mathbb R.}
$$
Each solution is considered on a connected interval in $\mathbb R\setminus\{-1,1\}$, where the original equation is defined.
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