# Deriving the logarithmic form of inverse hyperbolic cosecant

I am having trouble finding my mistake in deriving the logarithmic form of inverse hyperbolic cosecant function. Here is my work: $$y= \mathrm{csch} ^{-1} x \implies \mathrm {csch} \ y= x$$ $$\frac{2}{e^y-e^{-y}} = x$$ Multiplying the numerator and denominator by $e^y$ (this is possible since $e^y$ is never zero), $$\frac{2e^y}{e^{2y}-1} = x$$ Cross multiplying (this is possible since $y$ is never zero and thus $e^{2y}-1$ is never zero), $$xe^{2y}-x=2e^y$$ $$xe^{2y}-2e^y-x=0$$ $$e^y=\frac{2 \pm \sqrt{4+4x^2}}{2x}$$ Since $e^y \gt 0$ for all $y$, $$e^y=\frac{2 +\sqrt{4+4x^2}}{2x} = \frac{1 +\sqrt{1+x^2}}{x}$$ Hence, $$y= \ln \frac{1 +\sqrt{1+x^2}}{x}$$ However, we know that the logarithmic form of inverse hyperbolic cosecant function is $$\mathrm{csch} ^{-1} x = \ln( \frac{1}{x} + \frac{\sqrt{1+x^2}}{ \lvert x \rvert}).$$ My question is: where did I go wrong? My steps seem to be right, but I cannot figure out how I missed that absolute value symbol.

Please do NOT give another way of deriving the logarithmic form of inverse hyperbolic cosecant. I have already understood another way of deriving logarithmic form of inverse hyperbolic cosecant, a way which did in fact include that absolute value sign (I watched it on the following weblink: https://www.youtube.com/watch?v=278-6zKYDdE). I simply want to know why/how I missed that absolute value sign in my attempt to proof the formula.

The step from $$\tag{*} e^y=\frac{2 \pm \sqrt{4+4x^2}}{2x}$$ to $$e^y=\frac{2 +\sqrt{4+4x^2}}{2x} = \frac{1 +\sqrt{1+x^2}}{x} \, .$$ is only correct for $x > 0$. If $x< 0$ then the denominator in $(*)$ is negative and you have to choose the minus-sign in order to get a positive value for the expression.
Therefore $$e^y = \frac{1 + \text{sign}(x)\sqrt{1+x^2}}{x} = \frac 1x + \frac{\sqrt{1+x^2}}{|x|}$$