Q:

Show that the function $f(x) = \ln(x^2-2x+2)$ on $[1,\infty)$ has an inverse and find an expression for it.

I know that this function has an inverse since it is one to one. The reason it is one to one is because it's derivative is bigger than zero throughout the functions domain. What I got stuck at is the calculation when I tried to find the inverse function.
What I've done so far:
$y = \ln(x^2-2x+2)$
$e^y = e^{\ln(x^2-2x+2)}$ (cancels ln)
$e^y-2 = x^2 -2x$
Change $y$ and $x$
$e^x -2 = y^2 -2y $
$y(y-2) = e^x -2$
$y = (e^x-2)/(y-2)$
What do i do from here?

A:

Use the fact that\begin{align}e^y=x^2-2x+2&\iff e^y=(x-1)^2+1\\&\iff(x-1)^2=e^y-1\\&\iff x-1=\sqrt{e^y-1}.\end{align}

