Answer
$$1 - \ln 2$$
Work Step by Step
$$\eqalign{
& \iint\limits_R {\frac{1}{x}}dydx;\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \leqslant x \leqslant 2,\,\,\,\,\,\,0 \leqslant y \leqslant x - 1 \cr
& {\text{Replacing the limits for the region }}R \cr
& \int_1^2 {\int_0^{x - 1} {\frac{1}{x}dy} dx} \cr
& = \int_1^2 {\left[ {\int_0^{x - 1} {\frac{1}{x}dy} } \right]dx} \cr
& {\text{solve the inner integral treat }}x{\text{ as a constant}} \cr
& = \frac{1}{x}\int_0^{x - 1} {dy} \cr
& = \frac{1}{x}\left( y \right)_0^{x - 1} \cr
& {\text{evaluating the limits in the variable }}y \cr
& = \frac{1}{x}\left( {x - 1} \right) \cr
& = \frac{{x - 1}}{x} \cr
& = 1 - \frac{1}{x} \cr
& {\text{then}} \cr
& \cr
& \int_1^2 {\left[ {\int_0^{x - 1} {\frac{1}{x}dy} } \right]dx} = \int_1^2 {\left( {1 - \frac{1}{x}} \right)dx} \cr
& {\text{integrating}} \cr
& = \left[ {x - \ln \left| x \right|} \right]_1^2 \cr
& {\text{evaluating}} \cr
& = \left( {2 - \ln \left| 2 \right|} \right) - \left( {1 - \ln \left| 1 \right|} \right) \cr
& {\text{simplifying}} \cr
& = 2 - \ln 2 - 1 \cr
& = 1 - \ln 2 \cr} $$
