Solve logarithmic equation for $x$ to find the inverse of $f(x)= \ln(x+\sqrt{x^2+1})$ 
Let $f(x)= \ln(x+\sqrt{x^2+1})$. Find $f^{-1}(x)$.

Here is what I got so far: $y= \ln(x+\sqrt{x^2+1})$, rewrite as $x= \ln(y+\sqrt{y^2+1})$, 
then  $$e^x= y+\sqrt{y^2+1}$$ $$e^x-y= \sqrt{y^2+1}$$ $$  y^2+ e^{2x}-2(e^x)y= 1$$ 
So if   $e^x= a$, then $a^2-2ay-1= 0$

Let $y=\ln(x+\sqrt{x^2+1})$, then:
$$e^{y}=x+\sqrt{x^2+1}$$
$$e^{y}-x=\sqrt{x^2+1}$$
$$(e^{y}-x)^2=(\sqrt{x^2+1})^2$$
$$e^{2y}-2xe^{y}+x^2=x^2+1$$
$$e^{2y}-2xe^{y}=1$$
$$e^{2y}-1=2xe^{y}$$
$$x=\frac{e^{y}-e^{-y}}{2}$$
So $\displaystyle f^{-1}(y)=\frac{e^{y}-e^{-y}}{2}$.
