Answer
$$\frac{{15}}{{16}} + \ln 2$$
Work Step by Step
$$\eqalign{
& \int_1^4 {\left( {\frac{x}{8} + \frac{1}{{2x}}} \right)} dx \cr
& {\text{use the sum rule}} \cr
& = \int_1^4 {\frac{x}{8}} dx + \int_1^4 {\frac{1}{{2x}}} dx \cr
& = \frac{1}{8}\int_1^4 x dx + \frac{1}{2}\int_1^4 {\frac{1}{x}} dx \cr
& {\text{integrate using the power rule and }}\cr
&\int {\frac{1}{x}} dx = \ln \left| x \right| + C \cr
& = \frac{1}{8}\left( {\frac{{{x^2}}}{2}} \right)_1^4 + \frac{1}{2}\left( {\ln \left| x \right|} \right)_1^4 \cr
& = \left( {\frac{{{x^2}}}{{16}} + \frac{1}{2}\ln \left| x \right|} \right)_1^4 \cr
& {\text{use the fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = \left( {\frac{{{{\left( 4 \right)}^2}}}{{16}} + \frac{1}{2}\ln \left| 4 \right|} \right) - \left( {\frac{{{{\left( 1 \right)}^2}}}{{16}} + \frac{1}{2}\ln \left| 1 \right|} \right) \cr
& = \left( {1 + \frac{1}{2}\ln 4} \right) - \left( {\frac{1}{{16}} + 0} \right) \cr
& = 1 + \ln 2 - \frac{1}{{16}} \cr
& = \frac{{15}}{{16}} + \ln 2 \cr} $$
