Answer
$$e\ln 2e - e - \ln 2 + 1$$
Work Step by Step
$$\eqalign{
& \int_1^e {\ln 2x} dx \cr
& {\text{substitute }}u = \ln 2x,{\text{ }}du = \frac{1}{x}dx \cr
& dv = dx,{\text{ }}v = x \cr
& {\text{applying integration by parts}}{\text{, we have}} \cr
& \int_1^e {\ln 2x} dx = \left. {\left( {x\ln 2x} \right)} \right|_1^e - \int_1^e {\left( {x\left( {\frac{1}{x}} \right)} \right)dx} \cr
& \int_1^e {\ln 2x} dx = \left. {\left( {x\ln 2x} \right)} \right|_1^e - \int_1^e {dx} \cr
& {\text{integrating}} \cr
& \int_1^e {\ln 2x} dx = \left. {\left( {x\ln 2x} \right)} \right|_1^e - \left. {\left( x \right)} \right|_1^e \cr
& \int_1^e {\ln 2x} dx = \left. {\left( {x\ln 2x - x} \right)} \right|_1^e \cr
& {\text{evaluate limits}} \cr
& = \left( {e\ln 2e - e} \right) - \left( {1\ln 2 - 1} \right) \cr
& {\text{Simplify}} \cr
& = e\ln 2e - e - \ln 2 + 1 \cr} $$
