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Q.

$\int_{2}^{3} \frac{1}{x^{2} - x} d x =$

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a

$\ln \frac{1}{4}$

b

$\ln \frac{2}{3}$

c

$\ln \frac{4}{3}$

d

$\ln \frac{8}{3}$

answer is B.

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Detailed Solution

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$\int_{2}^{3} \frac{1}{x^{2} - x} d x$

$= \int_{2}^{3} \frac{1}{x (x - 1)} d x$

$= \int_{2}^{3} [\frac{- 1}{x} + \frac{1}{x - 1}] d x$

$= {(- \ln x + \ln (x - 1))}_{2}^{3}$

$= {(\ln \frac{x - 1}{x})}_{2}^{3}$

$= \ln (\frac{2}{3}) - \log \frac{1}{2} = \log \frac{2 / 3}{1 / 2} = \log \frac{4}{3}$

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