Answer
$\displaystyle \frac{2y-1}{8}\sqrt{6+4y-4y^{2}}-\frac{7}{8}\sin^{-1}\frac{1-2y}{\sqrt{7}}-\frac{1}{12}(6+4y-4y^{2})^{3/2}+C$
Work Step by Step
$6+4y-4y^{2}=6+1-1+4y-4y^{2}=7-(1-2y)^{2}$
If we substitute $u=(1-2y)$, then $du=-2dy$ and $y=\displaystyle \frac{1-u}{2}$,
and the integral becomes
$\displaystyle \int\frac{1-u}{2}\sqrt{(\sqrt{7})^{2}-u^{2}}\frac{du}{-2}=-\frac{1}{4}\int\sqrt{(\sqrt{7})^{2}-u^{2}}du+\frac{1}{4}\int u\sqrt{(\sqrt{7})^{2}-u^{2}}du$
For $I_{1}=\displaystyle \int\sqrt{(\sqrt{7})^{2}-u^{2}}du$ we use formula
30. $\displaystyle \int\sqrt{\mathrm{a}^{2}-u^{2}}du=\frac{u}{2}\sqrt{a^{2}-u^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{u}{a}+C_{1}$
For $I_{2}$, we substitute $\left[t=7-u^{2} , dt=-2udu\right]$
$I_{2}=\displaystyle -\frac{1}{2}\int\sqrt{t}dt=-\frac{1}{3}t^{3/2}+C_{2}=-\frac{1}{3}(7-u^{2})^{3/2}+C_{2}$
$=\displaystyle -\frac{1}{3}(6+4y-4y^{2})^{3/2}+C_{2}$
$\displaystyle \int y\sqrt{6+4y-4y^{2}}dy=-\frac{1}{4}I_{1}+\frac{1}{4}I_{2}$
$=-\displaystyle \frac{1}{4}(\frac{1-2y}{2}\sqrt{6+4y-4y^{2}}+\frac{7}{2}\sin^{-1}\frac{1-2y}{\sqrt{7}})+\frac{1}{4}(-\frac{1}{3}(6+4y-4y^{2})^{3/2})+C$
$=\displaystyle \frac{2y-1}{8}\sqrt{6+4y-4y^{2}}-\frac{7}{8}\sin^{-1}\frac{1-2y}{\sqrt{7}}-\frac{1}{12}(6+4y-4y^{2})^{3/2}+C$
