Answer
$$\frac{1}{3}\left( {{e^{15 + 5y}} - {e^{3 + 5y}}} \right)$$
Work Step by Step
$$\eqalign{
& \int_1^5 {{e^{3x + 5y}}} dx \cr
& {\text{the notation }}dx{\text{ indicates integration with respect to }}x,{\text{ }} \cr
& {\text{so we treat }}x{\text{ as variable and }}y{\text{ as a constant}}{\text{. then}} \cr
& \int_1^5 {{e^{3x + 5y}}} dx = \frac{1}{3}\int_1^5 {{e^{3x + 5y}}} \left( 3 \right)dx \cr
& {\text{using }}\int {{e^u}} du = {e^u} + C \cr
& = \frac{1}{3}\left( {{e^{3x + 5y}}} \right)_1^5 \cr
& {\text{evaluating the limits in the variable }}x \cr
& = \frac{1}{3}\left( {{e^{3\left( 5 \right) + 5y}} - {e^{3\left( 1 \right) + 5y}}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{1}{3}\left( {{e^{15 + 5y}} - {e^{3 + 5y}}} \right) \cr} $$
