## Thomas' Calculus 13th Edition

$$2\ln 10$$
\eqalign{ & \int_{1/10}^{10} {\frac{{{{\log }_{10}}\left( {10x} \right)}}{x}} dx \cr & {\text{using the property }}{\log _a}u = \frac{{\ln u}}{{\ln a}}{\text{ }}\left( {{\text{see example 7b}}} \right) \cr & \int_{1/10}^{10} {\frac{{{{\log }_{10}}\left( {10x} \right)}}{x}} dx = \int_{1/10}^{10} {\frac{{\ln 10x}}{{x\ln 10}}} dx \cr & {\text{use the product property for logarithms: }}\cr & \ln ab = \ln a + \ln b \cr & = \int_{1/10}^{10} {\frac{{\ln 10 + \ln x}}{{x\ln 10}}} dx \cr & {\text{distribute the numerator}} \cr & = \int_{1/10}^{10} {\left( {\frac{{\ln 10}}{{x\ln 10}} + \frac{{\ln x}}{{x\ln 10}}} \right)} dx \cr & {\text{simplifying}} \cr & = \int_{1/10}^{10} {\left( {\frac{1}{x} + \frac{{\ln x}}{{x\ln 10}}} \right)} dx \cr & {\text{split integrand}} \cr & = \int_{1/10}^{10} {\frac{1}{x}} dx + \frac{1}{{\ln 10}}\int_{1/10}^{10} {\ln x\left( {\frac{1}{x}} \right)} dx \cr & {\text{use }}\int {\frac{1}{x}dx = \ln \left| x \right| + C{\text{ and}}} {\text{ the power rule }}\int {{u^n}du = \frac{{{u^{n + 1}}}}{{n + 1}} + C}\cr & {\text{ note that }}u = \ln x \cr & = \left( {\ln \left| x \right|} \right)_{1/10}^{10} + \frac{1}{{\ln 10}}\left( {\frac{{{{\ln }^2}x}}{2}} \right)_{1/10}^{10} \cr & = \left( {\ln \left| x \right| + \frac{{{{\ln }^2}x}}{{2\ln 10}}} \right)_{1/10}^{10} \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 & = \left( {\ln 10 + \frac{{{{\ln }^2}10}}{{2\ln 10}}} \right) - \left( {\ln \left( {\frac{1}{{10}}} \right) + \frac{{{{\ln }^2}\left( {1/10} \right)}}{{2\ln 10}}} \right) \cr & {\text{simplifying}} \cr & = \left( {\ln 10 + \frac{{\ln 10}}{2}} \right) - \left( {\ln {{10}^{ - 1}} + \frac{{{{\left( {\ln {{10}^{ - 1}}} \right)}^2}}}{{2\ln 10}}} \right) \cr & = \left( {\frac{3}{2}\ln 10} \right) - \left( { - \ln 10 + \frac{{{{\left( {\ln 10} \right)}^2}}}{{2\ln 10}}} \right) \cr & = \left( {\frac{3}{2}\ln 10} \right) - \left( { - \ln 10 + \frac{{\ln 10}}{2}} \right) \cr & = \frac{3}{2}\ln 10 + \frac{1}{2}\ln 10 \cr & = 2\ln 10 \cr}