Answer
$ \displaystyle \frac{(4x^{3}+1)|4x^{3}+1|}{24}+C$
Work Step by Step
See table Shortcuts: Integrals of Expressions Involving $(ax+b)$
Target: $\displaystyle \int|ax+b|dx =\displaystyle \frac{1}{2a}(ax+b)|ax+b|+C$
$\displaystyle \mathrm{I}=\int x^{2}|4x^{3}+1|dx$
Substitute $\left[\begin{array}{lll}
u=x^{3} & & \\
du=3x^{2}dx, & \Rightarrow & x^{2}dx=\frac{du}{3}
\end{array}\right]$
$I=\displaystyle \int|4u+1|\frac{du}{3}=\frac{1}{3}\int|4u+1|du$
we have the target integral,$\quad a=4, b=1$
$=\displaystyle \frac{1}{3}\cdot\frac{1}{2(4)}(4u+1)|4u+1|+C$
bring back the variable $x$
=$ \displaystyle \frac{(4x^{3}+1)|4x^{3}+1|}{24}+C$
