$\int \sqrt{2 - 3 y - 4 y^{2}} d y is equal to$

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$\frac{1}{16} (8 y + 3) \sqrt{2 - 3 y - 4 y^{2}} + \frac{41}{64} \sin^{- 1} (\frac{8 y + 3}{\sqrt{41}}) + C$

$\frac{1}{8} (8 y + 3) \sqrt{2 - 3 y - 4 y^{2}} + \frac{41}{64} \sin^{- 1} (\frac{8 y + 3}{\sqrt{41}}) + C$

$\frac{1 (8 y + 3)}{16} \sqrt{2 - 3 y - 4 y^{2}} + \frac{41}{64} \sin^{- 1} (\frac{8 y + 3}{\sqrt{41}}) + C$

$\frac{1}{16} (8 y + 3) \sqrt{2 - 3 y - 4 y^{2}} - \frac{41}{64} \sin^{- 1} (\frac{8 y + 3}{\sqrt{41}}) + C$

answer is A.

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Detailed Solution

$I = \int \sqrt{2 - 3 y - 4 y^{2}} d y$

$\begin{array}{l} = \int \sqrt{(- 4) [y^{2} + \frac{3}{4} y - \frac{2}{4}]} d y \\ = 2 \int \sqrt{(-) [{(y + \frac{3}{8})}^{2} - \frac{9}{64} - \frac{2}{4}]} d y \end{array}$

$\begin{array}{l} = 2 \int \sqrt{({(\frac{41}{8})}^{2} - {(y + \frac{3}{8})}^{2}]} d y \\ = 2 \frac{(y + \frac{3}{8})}{2} \sqrt{{(\frac{41}{8})}^{2} - {(y + \frac{3}{8})}^{2}} + \frac{41}{128} \sin^{- 1} (\frac{8 y + 3}{\sqrt{41}}) + C \\ = \frac{1}{16} (8 y + 3) \sqrt{2 - 3 y - 4 y^{2}} + \frac{41}{64} \sin^{- 1} (\frac{8 y + 3}{\sqrt{41}}) + C \end{array}$