Answer
$$\frac{1}{2}\ln 2$$
Work Step by Step
$$\eqalign{
& \int_{1/2}^1 {\frac{1}{{2x}}} dx \cr
& {\text{drop out the constant}} \cr
& = \frac{1}{2}\int_{1/2}^1 {\frac{1}{x}} dx \cr
& {\text{integrate}} \cr
& = \frac{1}{2}\left[ {\ln \left| x \right|} \right]_{1/2}^1 \cr
& {\text{evaluate}} \cr
& = \frac{1}{2}\left[ {\ln \left| 1 \right| - \ln \left| {1/2} \right|} \right] \cr
& = \frac{1}{2}\ln 2 \cr} $$