Answer
$$\ln 10$$
Work Step by Step
$$\eqalign{
& \int_0^9 {\frac{{2{{\log }_{10}}\left( {x + 1} \right)}}{{x + 1}}} dx \cr
& {\text{using the property }}{\log _a}u = \frac{{\ln u}}{{\ln a}}{\text{ }}\left( {{\text{see example 7b}}} \right) \cr
& \int_0^9 {\frac{{2{{\log }_{10}}\left( {x + 1} \right)}}{{x + 1}}} dx = \int_0^9 {\frac{{2\ln \left( {x + 1} \right)}}{{\left( {x + 1} \right)\ln 10}}} dx \cr
& = \frac{2}{{\ln 10}}\int_0^9 {\ln \left( {x + 1} \right)\frac{1}{{\left( {x + 1} \right)}}} dx \cr
& {\text{integrate using the power rule }}\int {{u^n}du = \frac{{{u^{n + 1}}}}{{n + 1}} + C} \cr
& {\text{for this exercise, we can note that }}u = \ln \left( {x + 1} \right);{\text{ then}} \cr
& = \frac{2}{{\ln 10}}\left( {\frac{{{{\ln }^2}\left( {x + 1} \right)}}{2}} \right)_0^9 \cr
& = \frac{1}{{\ln 10}}\left( {{{\ln }^2}\left( {x + 1} \right)} \right)_0^9 \cr
& {\text{use fundamental theorem of calculus: }}\cr
& \int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\left( {{\text{see page 281}}} \right) \cr
& = \frac{1}{{\ln 10}}\left( {{{\ln }^2}\left( {9 + 1} \right) - {{\ln }^2}\left( {0 + 1} \right)} \right) \cr
& {\text{simplifying}} \cr
& = \frac{1}{{\ln 10}}\left( {{{\ln }^2}\left( {10} \right) - 0} \right) \cr
& = \ln 10 \cr} $$
