Answer
$ -4e^{x}+\displaystyle \frac{x|x|}{6}-\frac{x}{8}+C$
Work Step by Step
applying Sum and Difference Rules,
... $=\displaystyle \int-4e^{x}dx+\int\frac{1}{3}|x|dx-\int\frac{1}{8}dx$
... Constant Multiple Rule
$=-4\displaystyle \int e^{x}dx+\frac{1}{3}\int|x|dx-\frac{1}{8}\int 1dx$
... first integral: $\displaystyle \int e^{x}dx=e^{x}+C$,
... second: $\displaystyle \int|x|dx=\frac{x|x|}{2}+C$,
$... $third: $\displaystyle \int 1dx=x+C$,
$=-4e^{x}+\displaystyle \frac{1}{3}\cdot\frac{x|x|}{2}-\frac{1}{8}x+C$
$=-4e^{x}+\displaystyle \frac{x|x|}{6}-\frac{x}{8}+C$
