The function is f(x)=ln($x+\sqrt(x^2+1)$)
Note- The square root is over the whole $x^2+1$
$f(x) = \ln\left(x + \sqrt{x^2+1}\right)$
a) we know the domain of the square root function is $x \geq 0$
b) we know the domain of the natural log is $x > 0$
so
$x^2 + 1 \geq 0 \Rightarrow x \in \mathbb{R}$
$x + \sqrt{x^2 + 1} > 0$
this is also true for $x \in \mathbb{R}$
so the domain of $f(x)$ is $\mathbb{R}$