Answer
$$\frac{{{{\ln }^2}4}}{2}$$
Work Step by Step
$$\eqalign{
& \int_1^4 {\frac{{\ln 2{{\log }_2}x}}{x}} dx \cr
& {\text{use constant multiple rule}} \cr
& = \ln 2\int_1^4 {\frac{{{{\log }_{10}}x}}{x}} dx \cr
& {\text{using the property }}{\log _a}x = \frac{{\ln x}}{{\ln a}}{\text{ }}\left( {{\text{see example 7b}}} \right) \cr
& = \ln 2\int_1^4 {\frac{{\ln x}}{{x\ln 2}}} dx \cr
& = \int_1^4 {\frac{{\ln x}}{x}} dx \cr
& = \int_1^4 {\ln x\left( {\frac{1}{x}} \right)} dx \cr
& {\text{integrate using the power rule }}\int {{u^n}du = \frac{{{u^{n + 1}}}}{{n + 1}} + C} \cr
& {\text{for this exercise, we can note that }}u = \ln x{\text{; then:}} \cr
& = \left( {\frac{{{{\left( {\ln x} \right)}^2}}}{2}} \right)_1^4 \cr
& {\text{use fundamental theorem of calculus: }}\cr
& \int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\left( {{\text{see page 281}}} \right) \cr
& = \left( {\frac{{{{\left( {\ln 4} \right)}^2}}}{2} - \frac{{{{\left( {\ln 1} \right)}^2}}}{2}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{{{{\ln }^2}4}}{2} \cr} $$
