$\int_{0}^{1} \frac{1}{\sqrt{2 + 3 x}} d x =$

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$\frac{2}{3} (\sqrt{5} - \sqrt{2})$

$\frac{2}{3} (\sqrt{5} + \sqrt{2})$

$\frac{3}{5} (\sqrt{5} + \sqrt{2})$

$\frac{2}{3} (\sqrt{3} + \sqrt{2})$

answer is A.

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Detailed Solution

$\int_{0}^{1} \frac{1}{\sqrt{2 + 3 x}} d x = \int_{0}^{1} {(2 + 3 x)}^{- \frac{1}{2}} d x = {[\frac{{(2 + 3 x)}^{\frac{1}{2}}}{3 (\frac{1}{2})}]}_{0 `}^{1}$

$= \frac{2}{3} [{(2 + 3. I)}^{\frac{1}{2}} - {(2 + 3.0)}^{\frac{1}{2}}] = \frac{2}{3} [\sqrt{5} - \sqrt{2}]$

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