Thomas' Calculus 13th Edition

$$\frac{4}{{\ln 2}}$$
\eqalign{ & \int_1^4 {\frac{{{2^{\sqrt x }}}}{{\sqrt x }}} dx \cr & {\text{Use substitution:}}\cr & {\text{Let }}u = \sqrt x,{\text{ then }}du = \frac{1}{{2\sqrt x }}dx,\,\,\,\,\frac{1}{{\sqrt x }}xdx = 2du \cr & {\text{transforming the limits of integration}} \cr & \,\,\,\,\,\,{\text{When }}x = 1,{\text{ }}u = {\left( {\sqrt 1 } \right)^2} = 1 \cr & \,\,\,\,\,\,{\text{When }}x = 4,{\text{ }}u = {\left( {\sqrt 4 } \right)^2} = 2 \cr & {\text{write the integral in terms of }}u \cr & \int_1^4 {\frac{{{2^{\sqrt x }}}}{{\sqrt x }}} dx = \int_1^2 {{2^u}} \left( {2du} \right) \cr & = 2\int_1^2 {{2^u}} du \cr & {\text{integrate using }}\int {{a^u}} du = \frac{{{a^u}}}{{\ln a}} + C \cr & = 2\left( {\frac{{{2^u}}}{{\ln 2}}} \right)_1^2 \cr & = \frac{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{2}{{\ln 2}}\left( {{2^2} - {2^1}} \right) \cr & {\text{simplifying}} \cr & = \frac{2}{{\ln 2}}\left( {4 - 2} \right) \cr & = \frac{4}{{\ln 2}} \cr}