The inverse hyperbolic sine $\sin^{-1}(x) = \mathrm{arcsinh}(x)$ is an odd function. This can be proved by manipulating the expression $\mathrm{arcsinh}(-x) = y$ as shown here.
But how to prove it through this definition instead?
$$\mathrm{arcsinh}(x) = \log \left( x + \sqrt{x^2 + 1} \right)$$
My attempt:
I applied this rule, so obtaining
$$\mathrm{arcsinh}(-x) = \log \left( \sqrt{x^2 + 1} \right) + \log \left( 1 - \frac{x}{\sqrt{x^2 + 1}} \right)$$
This does not seem the expression of $\mathrm{arcsinh}(x)$ with opposite sign. How to proceed?
$$\text{arsinh}(x)+\text{arsinh}(-x)=\log(\sqrt{x^2+1}+x)+\log(\sqrt{x^2+1}-x)=\log(x^2+1-x^2).$$
You can conclude.