The domain of the inverse function is the domain of the original function.

∫t =(x+ 1)/(x- 1)=(x- 1+2)/(x- 1)= 1+2/(x- 1)

∵x & gt； 1

∴ t= 1+2/(x- 1)> 1

∴y = ln[(x+ 1)/(x- 1)]= lnt & gt； 0

That is, the value range of the original function is (0, +∞).

The x range of the inverse function is (0, +∞).