Show that \(arcosh (x) = \ln \left(x+\sqrt{(x^2-1)}\right)\,, \quad x\geqslant 1 \)

\begin{align} \end{align}

\begin{align} Let\ y & = arcosh(x) \\ x & = cosh(y)\\ x & = \frac{e^y +e^{-y}}{2} \\ 2x & = e^y + e^{-y}\\ 2xe^y & = e^{2y} + e^{0}\\ 2xe^y & = e^{2y}+ 1\\ 0 & = e^{2y}-2xe^y+1 & \\ 0 & = (e^y-x)^2 -x^2 + 1 \\ e^y & = x \pm \sqrt{(x^2-1)}\\ \implies y & = \ln \left(x \pm \sqrt{(x^2-1)}\right) \\ But\ for\ x \gt 1\,, \quad & \ln \left(x- \sqrt{(x^2-1)} \right)\ \lt 1,\ and\ arcosh \lt 0.\\ However,\ by\ definition,\ & arcosh(x)\ must\ be\ positive.\ Therefore \\ arcosh (x) & = \ln \left(x+\sqrt{(x^2-1)} \right) \end{align}