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. 
 A: Try to use $$(\sqrt{x^2+1}-x)(\sqrt{x^2+1}+x)=1$$
A: You are almost done : Simply note that $f(x)+ f(-x) = \ln(\sqrt {x^2 + 1} + x) 
 + \ln(\sqrt{x^2 + 1}-x) = \\ \ln((\sqrt {x^2 + 1} + x)(\sqrt{x^2 + 1}-x)) = \ln (x^2 + 1 - x^2) = \ln 1 = 0$. 
Hence, the function is an odd function.
A: $$\ln(-x+\sqrt{x^2+1})=\ln\left(\frac{1}{\sqrt{x^2+1}+x}\right)=-\ln(\sqrt{x^2+1}+x)$$
A: Note that $\sqrt{1+x^2}$ is even function.
Now, $$\int_{-x}^{x}\frac{1}{\sqrt{1+t^2}}dt=2\int_{0}^{x}\frac{1}{\sqrt{1+t^2}}dt=2\ln(x+\sqrt{1+x^2})$$
Again, $$\int_{-x}^{x}\frac{1}{\sqrt{1+t^2}}dt=2\int_{-x}^{0}\frac{1}{\sqrt{1+t^2}}dt=-2\ln(-x+\sqrt{1+x^2})$$
Hence $\ln(x+\sqrt{1+x^2})=\ln(-x+\sqrt{1+x^2})$ i.e. odd function.
A: $$ f(\sinh\theta) = \theta $$
and both $\theta$ and $\sinh\theta$ are odd functions, hence $f$ is an odd function as well ($f(x)=\text{arcsinh}(x)$).
A: 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}}] =$
$\ln ( \frac{1}{x + \sqrt{x^2+1}} )=$
$- \ln (x + \sqrt{x^2+1})$.
