Is $\ln(\sqrt{x^2 + 1} - x)$ an odd function?

Question

$f(x)$ is an odd function if $f(-x) = -f(x)$.

If $f(x) = \ln(\sqrt{x^2 + 1} - x)$, one can observe graphically that $f(-x) = -f(x)$.

Hence, $f(x)$ must be an odd function. However, WolframAlpha gives the following result:

Is WolframAlpha wrong, or have I made an error?

Answer 1

You are right, indeed as noticed, we have that

$$f(x)+f(-x)=\ln(\sqrt{x^2 + 1} - x) +\ln(\sqrt{x^2 + 1} + x)=\ln(x^2+1-x^2)=\ln 1=0 $$

and therefore

$$f(x)+f(-x)=0 \iff f(x)=-f(-x)$$

Answer 2

You're right that it is odd. To see why, rationalize the argument of $\ln$ by multiplying and dividing by $\displaystyle \sqrt{x^2+1} - x$.

As for posing the question directly to WolframAlpha, it's hard to tell what it is doing or why it might be returning the wrong information like it is since their code base is not open source. My best guess is it is somehow invoking the complex logarithm which can muck stuff up. WolframAlpha (and Mathematica more generally) is not great at very specific things, I guess you found one such thing.