Answer
\[ = 2\,\left( {e - 1} \right)\]
Work Step by Step
\[\begin{gathered}
\int_0^1 {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}\,dx} \hfill \\
\hfill \\
integration\,\,of\,\,indefinite\,\,integral \hfill \\
\hfill \\
\int_{}^{} {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}} \,dx\,\,\, = \,\,2{e^{\sqrt x }} + C \hfill \\
\hfill \\
use\,\,the\,\,Definition\,\,of\,\,improper\,\,integral\, \hfill \\
\hfill \\
\int_a^b {f\,\left( x \right)} \,dx\,\, = \,\,\mathop {\lim }\limits_{c \to {a^ + }} \int_c^b {f\,\left( x \right)} \,dx \hfill \\
\hfill \\
provided\,\,the\,\,\,limit\,\,exists \hfill \\
\hfill \\
\int_0^1 {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}\,dx} = \,\,\mathop {\lim }\limits_{c \to {0^ + }} \int_c^1 {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}\,dx} \hfill \\
\hfill \\
then \hfill \\
\hfill \\
= \,\mathop {\lim }\limits_{c \to {0^ + }} \,\,\,\,\left[ {2{e^{\sqrt x }}} \right]_c^1\,\, \hfill \\
\hfill \\
use\,\,the\,\,ftc \hfill \\
\hfill \\
= \,\mathop {\lim }\limits_{c \to {0^ + }} \,\,\,\,\left[ {2e - {e^0}} \right] \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= 2\,\left( {e - 1} \right) \hfill \\
\end{gathered} \]