If f(x)=log_e(x+√(1+x^2)) then f^-1(x) is equal to

\(y=\log_e(x+\sqrt{1+x^2})\)

so $e^y=x+\sqrt{1+x^2}$  ...(1)

as it is an odd function

$e^{-y}-x+\sqrt{1+x^2}$  ...(2)

so eq(1) - eq(2)

$e^y-e^{-y}=2x$

so $\frac{e^y-e^{-y}}{2}$

so $f^{-1}(x)=\frac{e^x-e^{-x}}{2}$