Answer
$$A = 2\ln 3 - 4\ln 2$$
Work Step by Step
$$\eqalign{
& {\text{From the graph we can note that the area is given by:}} \cr
& A = \int_1^2 {\left( {\frac{{x - 2}}{x}} \right)} dx + \int_2^3 {\left( {\frac{{x - 2}}{x}} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left[ {x - 2\ln \left| x \right|} \right]_1^2 - \left[ {x - 2\ln \left| x \right|} \right]_2^3 \cr
& {\text{Simplifying}} \cr
& A = \left[ {2 - 2\ln \left| 2 \right|} \right] - \left[ {1 - 2\ln \left| 1 \right|} \right] - \left[ {3 - 2\ln \left| 3 \right|} \right] + \left[ {2 - 2\ln \left| 2 \right|} \right] \cr
& A = 2 - 2\ln 2 - 1 - 3 + 2\ln 3 + 2 - 2\ln 2 \cr
& A = 2\ln 3 - 4\ln 2 \cr} $$
