Answer
$-\displaystyle \frac{1}{2}(e^{-x-1}-1)|e^{-x-1}-1|+C$
Work Step by Step
In the table "Shortcut RuIe: lntegrals of Expressions Involving $(\mathrm{a}x+b)$", our target is
$\displaystyle \int|ax+b|dx =\displaystyle \frac{1}{2a}(ax+b)|ax+b|+C$
Substitute:
$\left[\begin{array}{ll}
u=e^{-x-1}, & du=(e^{-x-1}\cdot(-1))dx \\
& e^{-x-1}dx=-du
\end{array}\right]$
$\displaystyle \int|e^{-x-1}-1|(e^{-x-1})dx= -\int|u-1|du$
we have our target integral, $a=1, b=-1$
$=-(\displaystyle \frac{1}{2(1)}(u-1)|u-1|+C$
bring back the variable $x$
= $-\displaystyle \frac{1}{2}(e^{-x-1}-1)|e^{-x-1}-1|+C$
