Answer
$e^{x-1}+C$
Work Step by Step
Shortcut formula
$:\displaystyle \qquad \int e^{ax+b}dx=\frac{1}{a}e^{ax+b}+C$
$\displaystyle \int e^{x-1}dx=\qquad\left[\begin{array}{l}
a=1\\
b=-1
\end{array}\right]$
Apply the formula
$=\displaystyle \frac{1}{1}e^{(1)x-1}+C$
= $e^{x-1}+C$
