Answer
$$\ln 2$$
Work Step by Step
$$\eqalign{
& \int_2^3 {\frac{{2{{\log }_2}\left( {x - 1} \right)}}{{x - 1}}} dx \cr
& {\text{using the property }}{\log _a}u = \frac{{\ln u}}{{\ln a}}{\text{ }}\left( {{\text{see example 7b}}} \right) \cr
& \int_2^3 {\frac{{2{{\log }_2}\left( {x - 1} \right)}}{{x - 1}}} dx = \int_2^3 {\frac{{2\ln \left( {x - 1} \right)}}{{\left( {x - 1} \right)\ln 2}}} dx \cr
& = \frac{2}{{\ln 2}}\int_2^3 {\frac{{\ln \left( {x - 1} \right)}}{{\left( {x - 1} \right)}}} dx \cr
& = \frac{2}{{\ln 2}}\int_2^3 {\ln \left( {x - 1} \right)\frac{1}{{\left( {x - 1} \right)}}} dx \cr
& {\text{integrate using the power rule }}\int {{u^n}du = \frac{{{u^{n + 1}}}}{{n + 1}} + C} \cr
& {\text{for this exercise we can note that }}u = \ln \left( {x - 1} \right);{\text{ then}} \cr
& = \frac{2}{{\ln 2}}\left( {\frac{{{{\ln }^2}\left( {x - 1} \right)}}{2}} \right)_2^3 \cr
& = \frac{1}{{\ln 2}}\left( {{{\ln }^2}\left( {x - 1} \right)} \right)_2^3 \cr
& {\text{use fundamental theorem of calculus: }}\cr
& \int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\left( {{\text{see page 281}}} \right) \cr
& = \frac{1}{{\ln 2}}\left( {{{\ln }^2}\left( {3 - 1} \right) - {{\ln }^2}\left( {2 - 1} \right)} \right) \cr
& {\text{simplifying}} \cr
& = \frac{1}{{\ln 2}}\left( {{{\ln }^2}2 - {{\ln }^2}\left( 1 \right)} \right) \cr
& = \frac{1}{{\ln 2}}\left( {{{\ln }^2}2} \right) \cr
& = \ln 2 \cr} $$
