Answer
$$3 + \ln 2$$
Work Step by Step
$$\eqalign{
& \int_1^4 {\frac{{2x + 1}}{{2x}}} dx \cr
& {\text{Distribute the numerator}} \cr
& = \int_1^4 {\left( {\frac{{2x}}{{2x}} + \frac{1}{{2x}}} \right)} dx \cr
& = \int_1^4 {\left( {1 + \frac{1}{{2x}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = \left[ {x + \frac{1}{2}\ln \left| x \right|} \right]_1^4 \cr
& = \left[ {4 + \frac{1}{2}\ln \left| 4 \right|} \right] - \left[ {1 + \frac{1}{2}\ln \left| 1 \right|} \right] \cr
& = 4 + \frac{1}{2}\ln 4 - 1 \cr
& = 3 + \ln {4^{1/2}} \cr
& = 3 + \ln 2 \cr} $$
