Answer
$3$
Work Step by Step
Applying th.7, let $u=g(y)=4y-y^{2}+4y^{3}+1 \quad$ (a continuous function).
Then, $g'(y)=4-2y+12y^{2}=12y^{2}-2y+4.$
Then, $\quad \left[\begin{array}{ll}
u=4y-y^{2}+4y^{3}+1 & du=g'(y)dy\\
& du=(12y^{2}-2y+4)dy \\
&
\end{array}\right]$,
and, the borders change to
$g(0)=0-0+0+1=1$ and $\quad g(1)=4-1+4+1=8$
$\displaystyle \int_{0}^{1}(4y-y^{2}+4y^{3}+1 )^{-2/3}(12y^{2}-2y+4)dy=\displaystyle \int_{1}^{8}u^{-2/3}du$
$=\displaystyle \left[ \frac{u^{-2/3+1}}{-2/3+1} \right]_{1}^{8}$
$=\displaystyle \left[ \frac{u^{1/3}}{1/3} \right]_{1}^{8}$
$=3\left[ u^{1/3} \right]_{1}^{8}$
$=3(8^{1/3}-1)$
$=3(2-1)$
$=3$
