Answer
$$\int^1_0(y^3+6y^2-12y+9)^{-1/2}(y^2+4y-4)dy=-\frac{2}{3}$$
Work Step by Step
$$A=\int^1_0(y^3+6y^2-12y+9)^{-1/2}(y^2+4y-4)dy$$
We set $u=y^3+6y^2-12y+9$, which means $$du=(3y^2+12y-12)dy=3(y^2+4y-4)dy$$ $$(y^2+4y-4)dy=\frac{1}{3}du$$
For $y=1$, we have $$u=1+6-12+9=4$$
For $y=0$, we have $$u=0+0-0+9=9$$
Therefore, $$A=\frac{1}{3}\int^4_9u^{-1/2}du$$ $$A=\frac{1}{3}\times2u^{1/2}\Big]^4_9=\frac{2}{3}\sqrt u\Big]^4_9$$ $$A=\frac{2}{3}(\sqrt4-\sqrt9)$$ $$A=\frac{2}{3}(2-3)=-\frac{2}{3}$$
