Answer
$$\frac{1}{3}\left( {{e^{2x + 3}} - {e^{2x - 3}}} \right)$$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^1 {{e^{2x + 3y}}} dy \cr
& {\text{the notation }}dy{\text{ indicates integration with respect to }}y,{\text{ so we treat }}y{\text{ as variable and}} \cr
& x{\text{ as a constant}}{\text{. then}} \cr
& \int_{ - 1}^1 {{e^{2x + 3y}}} dy = \frac{1}{3}\int_{ - 1}^1 {{e^{2x + 3y}}} \left( 3 \right)dy \cr
& {\text{using }}\int {{e^u}} du = {e^u} + C \cr
& = \frac{1}{3}\left( {{e^{2x + 3y}}} \right)_{ - 1}^1 \cr
& {\text{evaluating the limits in the variable }}x \cr
& = \frac{1}{3}\left( {{e^{2x + 3\left( 1 \right)}} - {e^{2x + 3\left( { - 1} \right)}}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{1}{3}\left( {{e^{2x + 3}} - {e^{2x - 3}}} \right) \cr} $$
