## Calculus (3rd Edition)

$$\frac{1}{4}(e^9-e).$$
Let $u=4x-3$, then $du=4dx$. Now, we have $$\int_1^3 e^{4x-3}dx =\frac{1}{4} \int_1^9 e^udu = \frac{1}{4} e^u|_1^9=\frac{1}{4}(e^9-e).$$