Too Hyperbolic?

$\begin{aligned} f(x) & = \ln \left(x+\sqrt{x^2-1}\right) \\ \implies f \left( \sqrt{\color{#3D99F6}\frac {1+\cosh x}2}\right) & = \ln \left( \sqrt {\color{#3D99F6} \frac {1+\cosh x}2} + \sqrt {{\color{#3D99F6}\frac {1+\cosh x}2} - 1} \right) & \small \color{#3D99F6} \text{Note 1: } \frac {1+\cosh x}2 = \cosh^2 \frac x2. \\ & = \ln \left( \sqrt {\color{#3D99F6} \cosh^2 \frac x2} + \sqrt {{\color{#3D99F6}\cosh^2 \frac x2} - 1} \right) \\ & = \ln \left(\cosh \frac x2 + \sqrt {\color{#3D99F6}\cosh^2 \frac x2 - 1} \right) & \small \color{#3D99F6} \text{Note 2: } \cosh^2 x - 1 = \sinh^2 x. \\ & = \ln \left(\cosh \frac x2 + \sqrt {\color{#3D99F6}\sinh^2 \frac x2} \right) \\ & = \ln \left({\color{#3D99F6}\cosh \frac x2 + \sinh \frac x2} \right) & \small \color{#3D99F6} \text{Note 3: } \cosh x + \sinh x = e^x. \\ & = \ln \left({\color{#3D99F6} e^\frac x2} \right) \\ & = \boxed{\dfrac x2} \end{aligned}$

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Notes:

$\begin{aligned} 1: \ \ \frac {1+\cosh x}2 & = \frac {1+\frac 12 \left(e^x + e^{-x} \right)}2 = \frac {e^x + 2 + e^{-x}}4 = \left(\frac {e^\frac x2 + e^{-\frac x2}}2 \right)^2 = \cosh^2 \frac x2 \end{aligned}$

$\begin{aligned} 2: \ \ \cosh^2 x - 1 & = \left(\frac {e^x + e^{-x}}2\right)^2 - 1 = \frac {e^{2x} + 2 + e^{-2x}-4}4 = \left(\frac {e^x - e^{-x}}2\right)^2 = \sinh^2 x \end{aligned}$

$\begin{aligned} 3: \ \ \cosh x + \sinh x & = \frac {e^x + e^{-x}}2 + \frac {e^x - e^{-x}}2 = e^x \end{aligned}$