Calculus with Applications (10th Edition)

$$\ln 20 - 1$$
\eqalign{ & \int_1^2 {\ln 5x} dx \cr & {\text{setting }}\,\,\,\,\,\,u = \ln 5x{\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 5x} dx = x\ln 5x - \int {x\left( {\frac{1}{x}dx} \right)} \cr & \int {\ln 5x} dx = x\ln 5x - \int {dx} \cr & {\text{integrate}} \cr & \int {\ln 5x} dx = x\ln 5x - x + C \cr & \cr & {\text{Now find the definite integral}} \cr & \int_1^2 {\ln 5x} dx = \left. {\left( {x\ln 5x - x} \right)} \right|_1^2 \cr & {\text{evaluating the limits}} \cr & = \left( {2\ln 5\left( 2 \right) - 2} \right) - \left( {\ln 5\left( 1 \right) - 1} \right) \cr & {\text{simplifying}} \cr & = 2\ln 10 - 2 - \ln 5 + 1 \cr & = 2\ln 10 - \ln 5 - 1 \cr & = \ln 100 - \ln 5 - 1 \cr & = \ln \left( {\frac{{100}}{5}} \right) - 1 \cr & = \ln 20 - 1 \cr}