Answer
$-e^{-x}+C$
Work Step by Step
Shortcut formula$:\displaystyle \qquad \int e^{ax+b}dx=\frac{1}{a}e^{ax+b}+C$
$\displaystyle \int e^{-x}dx=\qquad\left[\begin{array}{l}
a=-1\\
b=0
\end{array}\right]$
Apply the formula
$=\displaystyle \frac{1}{-1}e^{(-1)x+0}+C$
= $-e^{-x}+C$
