Answer
$$\frac{{{{\ln }^2}2}}{2}$$
Work Step by Step
$$\eqalign{
& \int_1^2 {\frac{{\ln x}}{x}} dx \cr
& = \int_1^2 {\ln x\left( {\frac{1}{x}} \right)} dx \cr
& {\text{use substitution}}{\text{. Let }}u = \ln x,{\text{ so that }}du = \frac{1}{x}dx \cr
& {\text{the new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = 2,{\text{ then }}u = \ln 2 \cr
& \,\,\,\,\,\,{\text{If }}x = 1,{\text{ then }}u = \ln 1 = 0 \cr
& {\text{Then}} \cr
& \int_1^2 {\ln x\left( {\frac{1}{x}} \right)} dx = \int_0^{\ln 2} {udu} \cr
& {\text{use }}\int_a^b {{u^n}} du = \left( {\frac{{{u^{n + 1}}}}{{n + 1}}} \right)_a^b \cr
& = \left( {\frac{{{u^2}}}{2}} \right)_0^{\ln 2} \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = \frac{{{{\left( {\ln 2} \right)}^2}}}{2} - \frac{{{0^2}}}{2} \cr
& = \frac{{{{\ln }^2}2}}{2} \cr} $$
