<s>[INST] 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 $a^2-2ay-1= 0$
 [/INST] here is a trick in this particular situation. that is to recognize 
$$(\sqrt{x^2 + 1} +x)(\sqrt{x^2 + 1} - x) = 1 $$ so that 
$$\ln(\sqrt{x^2 + 1} + x) = -\ln(\sqrt{x^2 + 1} - x)$$
now we can find the inverse function. suppose $$\ln(\sqrt{x^2 + 1} + x) = y,$$
then $$\ln(\sqrt{x^2 + 1} - x) = -y$$ exponentiating these two equations and subtracting gives you $$x = {e^y - e^{-y} \over 2} = \sinh y $$
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