Show that \(arsinh (x) = \ln (x+\sqrt{(x^2+1)}) \)

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\begin{align} Let\ y & = arsinh(x) \\ x & = sinh(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)}\\ But \quad \sqrt {(x^2+1)} \gt x \\ So\ the\ only\ valid\ solution\ is \\ e^y & = x + \sqrt{(x^2+1)}\\ \implies y & = \ln \left(x+\sqrt{(x^2+1)}\right) \\ arsinh (x) & = \ln \left(x+\sqrt{(x^2+1)} \right) \end{align}