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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?

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    $\begingroup$ $(x+\sqrt{x^2+1})(-x+\sqrt{x^2+1})=1$ $\endgroup$ – Lord Shark the Unknown Nov 24 '18 at 14:37
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    $\begingroup$ The rule pulls you away from the solution, don't use it. $\endgroup$ – Yves Daoust Nov 24 '18 at 14:39
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$$\text{arsinh}(x)+\text{arsinh}(-x)=\log(\sqrt{x^2+1}+x)+\log(\sqrt{x^2+1}-x)=\log(x^2+1-x^2).$$

You can conclude.

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