And where this solution is defined? $\DeclareMathOperator{\arctanh}{arctanh}$

BTW, when collecting MT I have seen this problem and in many papers integration was atrocious.

As long as $|y|<1$ we have $\int \frac{dy}{1-y^2}=\arctanh (y)$ (inverse hyperbolic function $\tanh$) and then

$y=\tanh (t)=\frac{{e^t}-e^{-t}}{e^{t}+e^{-t}}$ (similar to $\int \frac{dy}{1+y^2}=\arctan(y)$).