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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.

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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 $$ e^y = \frac{1 + \text{sign}(x)\sqrt{1+x^2}}{x} = \frac 1x + \frac{\sqrt{1+x^2}}{|x|} $$

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