Inverse function of function consisting of ln

Question

How do we find the inverse function of $f(x)=\ln(\frac{x^2+1}{x})$ over the domain $[1,\infty)$? Solving using traditional method,we get $f^{-1}(x)=\frac{e^x+\sqrt{(e^x)^2-4}}{2}$ and its conjugate. But a function should have a unique range,so i don't know which one of them to pick.

Answer 1

You have$$\ln\left(\frac{x^2+1}x\right)=y\iff x+\frac1x=e^y\iff x=\frac{e^y\pm\sqrt{e^{2y}-4}}2.$$ Now, note that$$\frac{e^y+\sqrt{e^{2y}-4}}2\times\frac{e^y-\sqrt{e^{2y}-4}}2=\frac{e^{2y}-(e^{2y}-4)}4=1.$$ So, since we have two numbers here whose product is $1$ and such that one of them ($\frac{e^y+\sqrt{e^{2y}-4}}2$) is greater than $0$, then both of them are greater than $0$ and actually, one of them is greater than or equal to $1$ and the other one is smaller than or equal to $1$. But we are assuming that $x\geqslant1$. So, you pick the solution $\frac{e^y+\sqrt{e^{2y}-4}}2$.