Answer
$$\int {{{\log }_b}xdx} = \frac{1}{{\ln b}}\left( {x\ln x - x} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {{{\log }_b}xdx} \cr
& {\text{Recall the logaritmic property lo}}{{\text{g}}_a}b = \frac{{\ln b}}{{\ln a}},{\text{ then}} \cr
& \int {{{\log }_b}xdx} = \int {\left( {\frac{{\ln x}}{{\ln b}}} \right)} dx \cr
& {\text{Pull out the constant }} \cr
& \int {{{\log }_b}xdx} = \frac{1}{{\ln b}}\int {\ln x} dx \cr
& {\text{Use the Integral of }}\ln x\left( {{\text{See page 519}}} \right):{\text{ }} \cr
& \int {\ln xdx} = x\ln x - x + C,{\text{ then}} \cr
& \int {{{\log }_b}xdx} = \frac{1}{{\ln b}}\left( {x\ln x - x} \right) + C \cr
& {\text{The statement has been proved}} \cr} $$
