Why is $\log(\sqrt{x^2+1}+x)$ odd?

If$$f(x) = \log(\sqrt{x^2+1}+x)$$ then $$f(-x) = \log \left(\sqrt{(-x)^2+1}-x\right)=$$ $$= \log \left((\sqrt{x^2+1}-x)\cdot\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}\right)=$$ $$= \log \left(\frac{1}{\sqrt{x^2+1}+x}\right)=- \log({\sqrt{x^2+1}+x})=-f(x)$$

Hint: $(\sqrt{x^2+1}+x)(\sqrt{x^2+1}-x) = 1$.

We have $f(-x)=\log \left(\frac{(\sqrt{x^2+1}-x)(\sqrt{x^2+1}+x)}{(\sqrt{x^2+1}+x)}\right)=\log\left(\frac{x^2+1-x^2}{\sqrt{x^2+1}+x}\right)=-\log(\sqrt{x^2+1}+x)=-f(x)$.