Question 380351
This is equivalent to finding the power series of ln x centered around x = 1. Note that all derivatives of ln x at x = 1 are equal to 1 or -1. Since we have the power series

{{{ln (x) = (x-1) - ((x-1)^2)/2 + ((x-1)^3)/3 - ...}}}

Adding one to all the x terms produces the given result.