Answer
$$\frac{1}{{\ln 2}}$$
Work Step by Step
$$\eqalign{
& \int_1^{\sqrt 2 } {x{2^{\left( {{x^2}} \right)}}dx} \cr
& {\text{Use substitution:}}\cr
& {\text{Let }}u = {x^2},{\text{ then }}du = 2xdx,\,\,\,\,\,xdx = \frac{1}{2}du \cr
& {\text{transforming the limits of integration}} \cr
& \,\,\,\,\,\,{\text{When }}x = 1,{\text{ }}u = {\left( 1 \right)^2} = 1 \cr
& \,\,\,\,\,\,{\text{When }}x = \sqrt 2,{\text{ }}u = {\left( {\sqrt 2 } \right)^2} = 2 \cr
& {\text{write the integral in terms of }}u \cr
& \int_1^{\sqrt 2 } {x{2^{\left( {{x^2}} \right)}}dx} = \int_1^2 {{2^u}} \left( {\frac{1}{2}du} \right) \cr
& = \frac{1}{2}\int_1^2 {{2^u}} du \cr
& {\text{integrate using }}\int {{a^u}} du = \frac{{{a^u}}}{{\ln a}} + C \cr
& = \frac{1}{2}\left( {\frac{{{2^u}}}{{\ln 2}}} \right)_1^2 \cr
& = \frac{1}{{2\ln 2}}\left( {{2^u}} \right)_1^2 \cr
& {\text{use fundamental theorem of calculus: }}\cr
& \int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = \frac{1}{{2\ln 2}}\left( {{2^2} - {2^1}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{1}{{2\ln 2}}\left( {4 - 2} \right) \cr
& = \frac{1}{{\ln 2}} \cr} $$
