Answer
$$\ln 4$$
Work Step by Step
$$\eqalign{
& \int_1^4 {\frac{{{{\log }_2}x}}{x}} dx \cr
& {\text{using the property }}{\log _a}x = \frac{{\ln x}}{{\ln a}}{\text{ }}\left( {{\text{see example 7b}}} \right) \cr
& \int {\frac{{{{\log }_2}x}}{x}} dx = \int {\frac{{\ln x}}{{x\ln 2}}} dx \cr
& = \frac{1}{{\ln 2}}\int {\frac{{\ln x}}{x}} dx \cr
& {\text{use substitution}}{\text{. Let }}u = \ln x,{\text{ then }}du = \frac{1}{x}dx \cr
& {\text{write the integral in terms of }}u \cr
& \frac{1}{{\ln 2}}\int {\frac{{\ln x}}{x}} dx = \frac{1}{{\ln 2}}\int u du \cr
& {\text{integrate using the power rule }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C} \cr
& = \frac{1}{{\ln 2}}\left( {\frac{{{u^2}}}{2}} \right) + C \cr
& {\text{simplifying}} \cr
& = \frac{1}{{2\ln 2}}{u^2} + C \cr
& {\text{write in terms of }}x{\text{ replace }}\ln x{\text{ for }}u \cr
& = \frac{{{{\ln }^2}x}}{{2\ln 2}} + C \cr
& {\text{Use the result with limits of integration for }}x \cr
& \int_1^4 {\frac{{{{\log }_2}x}}{x}} dx = \left[ {\frac{{{{\ln }^2}x}}{{2\ln 2}}} \right]_1^4 \cr
& {\text{use fundamental theorem of calculus: }}\cr
& \int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = \frac{{{{\ln }^2}4}}{{2\ln 2}} - \frac{{{{\ln }^2}1}}{{2\ln 2}} \cr
& {\text{simplifying}} \cr
& = \frac{{{{\ln }^2}4}}{{2\ln 2}} = \frac{{{{\ln }^2}4}}{{\ln {2^2}}} = \frac{{{{\ln }^2}4}}{{\ln 4}} \cr
& = \ln 4 \cr} $$
