The identity $\mathrm{arcsinh}\left(x\right)\=\ln \left(x\+\sqrt{x^2\+1}\right)$ can be established as follows.

Set $y\=\mathrm{arcsinh}\left(x\right)$ so $x\=\sinh \left(y\right)\=\frac{e^y-e^{-y}}{2}$. Set $u\=e^y$ so $e^{-y}\=1\/u$ and the equation $x\=\frac{u-1\/u}{2}$ becomes the quadratic (in $u$)

$u^2-2 x u-1\=0$

having solution $u\=x \pm \sqrt{x^2\+1}$. Because $u$ is an exponential, it must be positive, so

$e^y\=u\=x\+\sqrt{x^2\+1}$

from which it follows that $y\=\mathrm{arcsinh}\left(x\right)\=\ln \left(x\+\sqrt{x^2\+1}\right)$.$$