The inverse of the function$y={\log }_a(x+\sqrt{\left(x^2+1\right)};\left(a>0,a\ne 1\right)$  is

$\because y={\log }_a\left(x+\sqrt{\left(x^2+1\right)}\right)=f\left(x\right)\left(\text{say}\right)$

$\therefore x=f^{-1}\left(y\right)$

And $a^y=x+\sqrt{\left(x^2+1\right)}$     …….(i)

Or    $a^{-y}=\sqrt{\left(x^2+1\right)}-x$ ……..(ii)

Subtracting  Eq. (ii) from Eq.(i),then

$x=\frac{1}{2}\left(a^y-a^{-y}\right)$

$\Rightarrow f^{-1}\left(y\right)=\frac{1}{2}\left(a^y-a^{-y}\right)$

Hence, $f^{-1}\left(x\right)=\frac{1}{2}\left(a^x-a^{-x}\right)$