<s>[INST] 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?
 [/INST] since the definition of the $$\sqrt{x^2-1}$$ we get $$|x|>1$$ which is for $$x-\sqrt{x^2-1}$$ impossible
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