Answer
$${e^{2x - 1}} + C $$
Work Step by Step
$$\eqalign{
& \int {2{e^{2x - 1}}} dx \cr
& = 8\int {{e^{x + 1}}} dx \cr
& {\text{set }}u = 2x - 1{\text{ then }}\frac{{du}}{{dx}} = 2 \to \frac{{du}}{2} = dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {2{e^{2x - 1}}} dx = \int {2{e^u}} \left( {\frac{{du}}{2}} \right) \cr
& = \int {{e^u}} du \cr
& {\text{integrating}} \cr
& = {e^u} + C \cr
& {\text{replace }}2x - 1{\text{ for }}u \cr
& = {e^{2x - 1}} + C \cr} $$
