## Calculus (3rd Edition)

$$\int e^{9-2x}dx=-\frac{1}{2}\ e^{9-2x}+c.$$
Assume that $u=9-2x$, then $du=-2dx$ and hence $$\int e^{9-2x}dx=-\frac{1}{2}\int e^udu=-\frac{1}{2}\ e^u+c=-\frac{1}{2}\ e^{9-2x}+c.$$