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