Answer
$\displaystyle \frac{(4x^{2}+1)|4x^{2}+1|}{16}+C$
Work Step by Step
$I=\displaystyle \int x|4x^{2}-1|dx$
See table Shortcuts: Integrals of Expressions Involving $(ax+b)$
Target: $\displaystyle \quad\int|ax+b|dx =\displaystyle \frac{1}{2a}(ax+b)|ax+b|+C$
Substitute $\left[\begin{array}{lll}
u=x^{2} & & \\
du=2xdx, & \Rightarrow & xdx=\frac{du}{2}
\end{array}\right]$
$I=\displaystyle \int|4u+1|\frac{du}{2}=\frac{1}{2}\int|4u+1|du$
we have the target integral,$\quad a=4, b=1$
$=\displaystyle \frac{1}{2}\cdot\frac{1}{2(4)}(4u+1)|4u+1|+C$
bring back the variable $x$
= $\displaystyle \frac{(4x^{2}+1)|4x^{2}+1|}{16}+C$
