Answer
$$\frac{{{{\ln }^2}x}}{{2\ln 10}} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\log }_{10}}x}}{x}} dx \cr
& {\text{using the property }}{\log _a}x = \frac{{\ln x}}{{\ln a}}{\text{ }}\left( {{\text{see example 7b}}} \right) \cr
& \int {\frac{{{{\log }_{10}}x}}{x}} dx = \int {\frac{{\ln x}}{{x\ln 10}}} dx \cr
& = \frac{1}{{\ln 10}}\int {\frac{{\ln x}}{x}} dx \cr
& {\text{use substitution}}{\text{. Let }}u = \ln x,{\text{ then }}du = \frac{1}{x}dx \cr
& {\text{write the integral in terms of }}u \cr
& \frac{1}{{\ln 10}}\int {\frac{{\ln x}}{x}} dx = \frac{1}{{\ln 10}}\int u du \cr
& {\text{integrate using the power rule }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C} \cr
& = \frac{1}{{\ln 10}}\left( {\frac{{{u^2}}}{2}} \right) + C \cr
& {\text{simplifying}} \cr
& = \frac{1}{{2\ln 10}}{u^2} + C \cr
& {\text{write in terms of }}x{\text{; replace }}\ln x{\text{ for }}u \cr
& = \frac{{{{\ln }^2}x}}{{2\ln 10}} + C \cr} $$
