Show that $f(x)$ is an Odd Function 
Show that $$f(x) = \ln \left(x+\sqrt{x^2+1}\right)$$ is an odd function.

My attempt:
$$f(-x)=\ln\left(-x+\sqrt{(-x)^2+1}\right)=\ln\left(-x+\sqrt{x^2+1}\right).$$
How should I proceed? I know that if $f(-x)=-f(x)$, the function is odd. 

One more:
$f(x) = \ln (x + \sqrt{x^2+1})$, $x \in \mathbb{R}.$
$\star)$  $f(-x) = \ln(-x + \sqrt{x^2 +1}) =$
$\ln (\frac{1}{x+\sqrt{x^2+1}} )=$
$- \ln (x + \sqrt{x^2+1})$. 
Combining:
$f(-x) = - f(x)$.
Used : 
$ \ln [( -x +\sqrt{x^2+1}) \frac{x + \sqrt{x^2 +1}}{x+\sqrt{x^2+1}}] =$
$\ln ( \frac{1}{x + \sqrt{x^2+1}} )=$
$- \ln (x + \sqrt{x^2+1})$.
