Deriving $\cosh^{-1}{x}=\ln\left(x+\sqrt{x^2-1}\right)$ Let $y=\cosh^{-1}{x}$. Then, $x=\cosh{y}=\frac{1}{2}\left(e^y+e^{-y}\right)$. Multiplying by $2e^y$, we get $2xe^y=e^{2y}+1$. Solving $e^{2y}-2xe^y+1=0$ by the quadratic formula, we have $e^y=\frac{2x\pm\sqrt{4x^2-4}}{2}=x\pm\sqrt{x^2-1}$. We find that both roots are possible. Thus, $y=\ln\left(x\pm\sqrt{x^2-1}\right)$. Why did we fail to eliminate the minus?

The reason why we can dispose of the minus sign is that, firstly, $$(x+\sqrt{x^2-1})^{-1}=(x-\sqrt{x^2-1})$$
Therefore $$\ln(x\pm \sqrt{x^2-1})=\ln(x+\sqrt{x^2-1})^{\pm1}=\pm\ln(x+\sqrt{x^2-1})$$
But $\operatorname{arcosh}(x)$ is defined as non-negative, so the minus sign is not required
