∫23(3x+23x−2)2+(3x−23x+2)2−2 dx is equal to

(3x+23x−2)2+(3x−23x+2)2−2=(3x+23x−2−3x−23x+2)2∫23(3x+23x−2)2+(3x−23x+2)2−2 dx⇒∫23(3x+23x−2−3x−23x+2)dx=∫2324x9x2−4dx=43{log|9x2−4|}23=43log(7732)

∫23(3x+23x−2)2+(3x−23x+2)2−2 dx is equal to

(3x+23x−2)2+(3x−23x+2)2−2=(3x+23x−2−3x−23x+2)2∫23(3x+23x−2)2+(3x−23x+2)2−2 dx⇒∫23(3x+23x−2−3x−23x+2)dx=∫2324x9x2−4dx=43{log|9x2−4|}23=43log(7732)

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$\int_{2}^{3} \sqrt{{(\frac{3 x + 2}{3 x - 2})}^{2} + {(\frac{3 x - 2}{3 x + 2})}^{2} - 2} d x$ is equal to

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$3 \log_{e} (\frac{8}{3})$

$\log_{e} (\frac{3}{8})$

$\frac{4}{3} \log (\frac{77}{32})$

$\log (\frac{9}{64})$

answer is C.

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Detailed Solution

${(\frac{3 x + 2}{3 x - 2})}^{2} + {(\frac{3 x - 2}{3 x + 2})}^{2} - 2 = {(\frac{3 x + 2}{3 x - 2} - \frac{3 x - 2}{3 x + 2})}^{2}$

$\int_{2}^{3} \sqrt{{(\frac{3 x + 2}{3 x - 2})}^{2} + {(\frac{3 x - 2}{3 x + 2})}^{2} - 2} d x$

$\Rightarrow \int_{2}^{3} (\frac{3 x + 2}{3 x - 2} - \frac{3 x - 2}{3 x + 2}) d x = \int_{2}^{3} \frac{24 x}{9 x^{2} - 4} d x = \frac{4}{3} {\log | 9 x^{2} - 4 |}_{2}^{3} = \frac{4}{3} \log (\frac{77}{32})$

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