Proving an Odd Function: ln(x+√(1+x²))

[Yankel]

Hello,

How do I prove that this function:

\[f(x)=ln(x+\sqrt{1+x^{2}})\]

is an odd function ?

It is clear to me that it is odd, I tried the usual way of checking if a function is even or odd, by looking at f(-x), but it just got me to where x is changing sign while the square root isn't, it there a logarithms rule I am missing here ?

Thank you !

 

[Siron]

$$f(-x) = \ln(-x+\sqrt{1+(-x)^2}) = \ln(-x+\sqrt{1+x^2})$$
Let us compute $-f(x)$:
$$-f(x) = - \ln(x+\sqrt{1+x^2}) = \ln\left(\frac{1}{x+\sqrt{1+x^2}}\right) = \ln\left[\frac{x-\sqrt{1+x^2}}{(x+\sqrt{1+x^2})(x-\sqrt{1+x^2})}\right]$$
$$= \ln\left(-x+\sqrt{1+x^2}\right)$$

Hence, $f(x)$ is odd.