Answer
$$\frac{1}{{\ln 2}}$$
Work Step by Step
$$\eqalign{
& \int_1^e {{x^{\left( {\ln 2} \right) - 1}}} dx \cr
& {\text{integrate}}{\text{, apply }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C,{\text{ }}where{\text{ }}n \ne - 1.{\text{ }}S{\text{o}}{\text{,}}} \cr
& = \left( {\frac{{{x^{\left( {\ln 2} \right) - 1 + 1}}}}{{\left( {\ln 2} \right) - 1 + 1}}} \right)_0^3 \cr
& {\text{simplifying}} \cr
& = \left( {\frac{{{x^{\ln 2}}}}{{\ln 2}}} \right)_1^e \cr
& = \frac{1}{{\ln 2}}\left( {{x^{\ln 2}}} \right)_1^e \cr
& {\text{use fundamental theorem of calculus: }}\cr
& \int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = \frac{{{e^{\ln 2}} - {1^{\ln 2}}}}{{\ln 2}} \cr
& = \frac{{2 - 1}}{{\ln 2}} \cr
& = \frac{1}{{\ln 2}} \cr} $$
