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I would like to know why this function can be written as it below.

And what is the h of hx in the cosh()?

$cosh(hx)= \left(\frac{x+ \frac{1}{x}}{2}\right)$

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    I think your "h" is supposed to be $ln(x)$. Note that $e^{\ln(x)}=x$. – 2017-01-11
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    It may be worthwhile to note that $hx$ and $ln(x)$ are somewhat typographically similar. Also, I have seen $ln(x)$ written as $ln x$ without parentheses making it more confusing. – 2017-01-11
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    @user71352 that's a good observation. +1 – 2017-01-11

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I would say that $hx=\log(x)$ (if you consider $h$ to be a function)

If you consider $h$ as number, then $h\cdot x=\log(x)$, so $h=\frac{\log(x)}{x}$