Answer
$$9\ln 27 - 8 - \ln 3$$
Work Step by Step
$$\eqalign{
& \int_1^9 {\ln 3x} dx \cr
& {\text{setting }}\,\,\,\,\,\,u = \ln 3x{\text{ then }}du = \frac{1}{x}dx\,\,\,\,\,\,\,\,\,\, \cr
& {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = dx{\text{ then }}v = x \cr
& {\text{Substituting these values into the formula for integration by parts}} \cr
& \int u dv = uv - \int {vdu} \cr
& \int {\ln 3x} dx = x\ln 3x - \int {x\left( {\frac{1}{x}dx} \right)} \cr
& \int {\ln 3x} dx = x\ln 3x - \int {dx} \cr
& {\text{integrate}} \cr
& \int {\ln 3x} dx = x\ln 3x - x + C \cr
& \cr
& {\text{Now find the definite integral}} \cr
& \int_1^9 {\ln 3x} dx = \left. {\left( {x\ln 3x - x} \right)} \right|_1^9 \cr
& {\text{evaluating the limits}} \cr
& = \left( {9\ln 3\left( 9 \right) - 9} \right) - \left( {\ln 3\left( 1 \right) - 1} \right) \cr
& {\text{simplifying}} \cr
& = 9\ln 27 - 9 - \ln 3 + 1 \cr
& = 9\ln 27 - 8 - \ln 3 \cr} $$
