Answer
$$8{e^{x + 1}} + C $$
Work Step by Step
$$\eqalign{
& \int {8{e^{x + 1}}} dx \cr
& = 8\int {{e^{x + 1}}} dx \cr
& {\text{set }}u = x + 1{\text{ then }}\frac{{du}}{{dx}} = 1 \to du = dx \cr
& {\text{write the integrand in terms of }}u \cr
& 8\int {{e^{x + 1}}} dx = 8\int {{e^u}} du \cr
& {\text{integrating}} \cr
& = 8{e^u} + C \cr
& {\text{replace }}x + 1{\text{ for }}u \cr
& = 8{e^{x + 1}} + C \cr} $$
