Answer
$-\displaystyle \frac{2}{3}$
Work Step by Step
Applying th.7, let $u=g(y)=y^{3}+6y^{2}-12y+9 \quad$ (a continuous function).
Then, $g'(y)=3y^{2}+12y-12=3(y^{2}+4y-4).$
Then, $\quad \left[\begin{array}{ll}
u=y^{3}+6y^{2}-12y+9 & du=g'(y)dy\\
& du=3(y^{2}+4y-4)dy \\
& (y^{2}+4y-4)d=\dfrac{du}{3}
\end{array}\right]$,
and, the borders change to
$g(0)=0+0-0+9=9$ and $\quad g(1)=1+6-12+9=4$
$\displaystyle \int_{0}^{1}(y^{3}+6y^{2}-12y+9 )^{-1/2}(y^{2}+4y-4)dy=\displaystyle \int_{9}^{4}u^{-1/2}(\frac{du}{3})$
$\displaystyle =\frac{1}{3}\int_{9}^{4}u^{-1/2}du$
$=\displaystyle \frac{1}{3}\left[ \frac{u^{-1/2+1}}{-1/2+1} \right]_{9}^{4}$
$=\displaystyle \frac{1}{3}\left[ \frac{u^{1/2}}{1/2} \right]_{9}^{4}$
$=\displaystyle \frac{2}{3}\left[ u^{1/2} \right]_{9}^{4}$
$=\displaystyle \frac{2}{3}(2-3)$
$=-\displaystyle \frac{2}{3}$
