#### Answer

$$\frac{{{3^{1 + \ln 2}} - 1}}{{\ln 3}}$$

#### Work Step by Step

$$\eqalign{
& \int_1^{2e} {\frac{{{3^{\ln x}}}}{x}} dx \cr
& {\text{substitute }}u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr
& {\text{express the limits in terms of }}u \cr
& x = 2e{\text{ implies }}u = \ln \left( {2e} \right) = 1 + \ln 2 \cr
& x = 1{\text{ implies }}u = \ln \left( 1 \right) = 0 \cr
& {\text{the entire integration is carried out as follows}} \cr
& \int_1^{2e} {\frac{{{3^{\ln x}}}}{x}} dx = \int_0^{1 + \ln 2} {{3^u}du} \cr
& {\text{find the antiderivative}} \cr
& = \left. {\left( {\frac{{{3^u}}}{{\ln 3}}} \right)} \right|_0^{1 + \ln 2} \cr
& {\text{evaluate limits}} \cr
& = \frac{{{3^{1 + \ln 2}}}}{{\ln 3}} - \frac{{{3^0}}}{{\ln 3}} \cr
& {\text{simplify}} \cr
& = \frac{{{3^{1 + \ln 2}} - 1}}{{\ln 3}} \cr} $$