Answer
$$2 - \frac{2}{{\sqrt e }}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{{dx}}{{\sqrt {{e^x}} }}} \cr
& {\text{radical property }}\sqrt {{a^m}} = {a^{m/2}} \cr
& = \int_0^1 {\frac{{dx}}{{{e^{x/2}}}}} \cr
& {\text{write with negative exponent}} \cr
& = \int_0^1 {{e^{ - x/2}}} dx \cr
& {\text{use }}\int {{e^{ - ax}}dx = - \frac{1}{a}{e^{ - ax}} + C} \cr
& = \left[ {\frac{1}{{ - 1/2}}{e^{ - x/2}}} \right]_0^1 \cr
& = \left[ { - \frac{2}{{{e^{x/2}}}}} \right]_0^1 \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = - \frac{2}{{{e^{1/2}}}} + \frac{2}{{{e^{0/2}}}} \cr
& {\text{simplify}} \cr
& = 2 - \frac{2}{{{e^{1/2}}}} \cr
& = 2 - \frac{2}{{\sqrt e }} \cr} $$
