The inverse function of

$\frac{1}{4}\left(\log _{8} \mathrm{e}\right) \log _{\mathrm{e}}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)$

$\frac{1}{4} \log _{\mathrm{e}}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)$

$\frac{1}{4}\left(\log _{8} e\right) \log _{e}\left(\frac{1+x}{1-x}\right)$

$\frac{1}{4} \log _{\mathrm{e}}\left(\frac{1+x}{1-x}\right)$