Answer
$${\ln ^2}2$$
Work Step by Step
$$\eqalign{
& \int_1^2 {\frac{{2\ln x}}{x}} dx \cr
& {\text{Use substitution:}}\cr
& {\text{Let }}u = \ln x,{\text{ so that }}du = \frac{1}{x}dx \cr
& {\text{The new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = 2,{\text{ }}u = \ln \left( 2 \right) \cr
& \,\,\,\,\,\,{\text{If }}x = 1,{\text{ }}u = \ln \left( 1 \right) = 0 \cr
& {\text{write the integral in terms of }}u \cr
& \int_1^2 {\frac{{2\ln x}}{x}} dx = 2\int_0^{\ln 2} {udu} \cr
& {\text{integrate by using the power rule}} \cr
& = 2\left( {\frac{{{u^2}}}{2}} \right)_0^{\ln 2} \cr
& = \left( {{u^2}} \right)_0^{\ln 2} \cr
& {\text{use the fundamental theorem of calculus: }}\cr
& \int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\left( {{\text{see page 281}}} \right) \cr
& = {\left( {\ln 2} \right)^2} - {\left( 0 \right)^2} \cr
& = {\ln ^2}2 \cr} $$
