Chapter 7.1, Problem 49E

Integrals of Functions Involving Absolute Values In Exercises 45–52, use integration by parts to evaluate the given integral using the following integral formulas where necessary. (You have seen some of these before; all can be checked by differentiating.)

Integral Formula	Shortcut Version
$\begin{array}{l}{\displaystyle \int \frac{\left|x\right|}{x}dx=\left|x\right|+C}\\ \text{ Because }\frac{d}{dx}\left|x\right|=\frac{\left|x\right|}{x}\end{array}$	${\displaystyle \int \frac{\left|ax+b\right|}{ax+b}dx=\frac{1}{a}\left|ax+b\right|+C}$
${\displaystyle \int \left|x\right|dx=\frac{1}{2}x\left|x\right|+C}$	${\displaystyle \int \left|ax+b\right|dx=\frac{1}{2a}\left(ax+b\right)\left|ax+b\right|+C}$
${\displaystyle \int x\left|x\right|dx=\frac{1}{3}{x}^2\left|x\right|+C}$	${\displaystyle \int \left(ax+b\right)\left|ax+b\right|dx=\frac{1}{3a}{\left(ax+b\right)}^2\left|ax+b\right|+C}$
${\displaystyle \int x^2\left|x\right|dx=\frac{1}{4}{x}^3\left|x\right|+C}$	${\displaystyle \int {\left(ax+b\right)}^2\left|ax+b\right|dx=\frac{1}{4a}{\left(ax+b\right)}^3\left|ax+b\right|+C}$

${\displaystyle \int 2x^2\left|-x+4\right|dx}$