int_{2}^{3} frac{dx}{x^2 - x} = | vedclass

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(C) Let $I = \int_{2}^{3} \frac{dx}{x(x - 1)}$.Using partial fractions,we can write $\frac{1}{x(x - 1)} = \frac{1}{x - 1} - \frac{1}{x}$.Therefore,$I

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$\log(4/3)$

Vedclass · Answer

(C) Let $I = \int_{2}^{3} \frac{dx}{x(x - 1)}$.Using partial fractions,we can write $\frac{1}{x(x - 1)} = \frac{1}{x - 1} - \frac{1}{x}$.Therefore,$I = \int_{2}^{3} \left( \frac{1}{x - 1} - \frac{1}{x} \right) dx$.Integrating term by term,we get $I = [\log|x - 1| - \log|x|]_{2}^{3}$.Using the property $\log a - \log b = \log(a/b)$,we have $I = [\log|\frac{x - 1}{x}|]_{2}^{3}$.Substituting the limits: $I = \log|\frac{3 - 1}{3}| - \log|\frac{2 - 1}{2}|$.$I = \log(2/3) - \log(1/2) = \log(\frac{2/3}{1/2}) = \log(4/3)$.

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

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