Chapter 13 - Section 13.2 - Substitution - Exercises - Page 971: 56

See below.

$\displaystyle \int e^{ax+b}dx=\qquad \left[\begin{array}{ll} u=ax+b, & du=adx, \\ & dx=\frac{1}{a}du \end{array}\right]$ $=\displaystyle \frac{1}{a}\int e^{u}du+C$ $=\displaystyle \frac{1}{a}e^{u}+C$ bring back the variable $x$ = $\displaystyle \frac{1}{a}e^{ax+b}+C$