<s>[INST] Prove that $\ln(x+\sqrt{x^2 + 4}) - \ln2$ is odd I know that a function is odd when
$$f(-x) = -f(x)$$
Therefore I can say that if for a function $$-f(x) + f(x) = f(-x) + f(x) = 0$$
Then the function is odd!
I tried to use this trick to prove that $f(x) = \ln\left(x+\sqrt{x^2 + 4}\right) - \ln2$ is odd.
However, I would want to prove directly that $$f(-x) = -f(x)$$
In other words, I want to solve $$\ln\left(-x+\sqrt{(-x)^2 + 4}\right) - \ln2$$
and to come at the end to this:
$$-\ln\left(x+\sqrt{x^2 + 4}\right) + \ln2$$
This was my approach:
$$\ln\left(-x+\sqrt{(-x)^2 + 4}\right) - \ln2$$
$$\ln\left(-x+\sqrt{x^2 + 4}\right) - \ln2$$
$$\ln\left(\frac{-x+\sqrt{x^2 + 4}}{2}\right)$$
$$\ln\left(\left(\frac{2}{-x+\sqrt{x^2 + 4}}\right)^{-1}\right)$$
$$-\ln\left(\frac{2}{-x+\sqrt{x^2 + 4}}\right)$$

Here I got stuck. I want to get to $-\ln\left(x+\sqrt{x^2 + 4}\right) + \ln2$ but if I use $\ln\left(\frac ab\right) = \ln a - \ln b$ then I will get back to $f(-x)$ and not to $-f(x)$.
Any help?
 [/INST] A faster approach. If we set $f(x)=\log\left(\frac{x+\sqrt{x^2+4}}{2}\right)$ and $x=2\sinh\theta$ we have 
$$f(2\sinh\theta) = \log\left(\frac{2\sinh\theta+2\cosh\theta}{2}\right) = \log(e^\theta) = \theta $$
and since $\theta\mapsto 2\sinh\theta$ is a bijective odd function from $\mathbb{R}$ to $\mathbb{R}$,
$$ f(-x) = -f(x) $$
readily follows. That also implies $f(x) = \text{arcsinh}\left(\frac{x}{2}\right)$.
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