Answer
$cosh^{-1}~x = ln(x+\sqrt{x^2-1})~~~~~~~$ for $~~x \geq 1$
Work Step by Step
Let $y = cosh^{-1}~x$
Then:
$x = cosh~y = \frac{e^y+e^{-y}}{2}$
$e^y-2x+e^{-y} = 0$
$e^{2y}-2xe^y+1 = 0$
We can use the quadratic formula:
$e^y = \frac{2x\pm\sqrt{4x^2-4}}{2}$
$e^y = x\pm\sqrt{x^2-1}$
This will lead to two solutions, but by convention, we can use $x+\sqrt{x^2-1}$ as the solution.
$e^y = x+\sqrt{x^2-1}$
$y = ln(x+\sqrt{x^2-1})$
$cosh^{-1}~x = ln(x+\sqrt{x^2-1})$
Note that $x\geq 1$ because of the term $\sqrt{x^2-1}$ and because it is required that $x+\sqrt{x^2-1} \gt 0$