TITLE: Why is $\log(\sqrt{x^2+1}+x)$ odd?
QUESTION [42 upvotes]: $$f(x) = \log(\sqrt{x^2+1}+x)$$
I can't figure out, why this function is odd. I mean, of course, its graph shows, it's odd, but when I investigated $f(-x)$, I couldn't find way to $-\log(\sqrt{x^2+1}+x)$.

REPLY [4 votes]: Note by differentiating that
$$f(x)=\int_0^x \frac{dt}{\sqrt{t^2+1}}.$$
The result now follows from the fact that $\dfrac{1}{\sqrt{t^2+1}}$ is even.