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

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(C) We are given the integral $I = \int_{2}^{3} \frac{dx}{x^{2}+x}$.First,we factor the denominator: $x^{2}+x = x(x+1)$.Using partial fractions,we can

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$\log \left(\frac{9}{8}\right)$

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(C) We are given the integral $I = \int_{2}^{3} \frac{dx}{x^{2}+x}$.First,we factor the denominator: $x^{2}+x = x(x+1)$.Using partial fractions,we can write $\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}$.Solving for $A$ and $B$,we get $1 = A(x+1) + Bx$. Setting $x=0$ gives $A=1$,and setting $x=-1$ gives $B=-1$.Thus,$\frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1}$.Now,integrate: $I = \int_{2}^{3} \left(\frac{1}{x} - \frac{1}{x+1}ight) dx = [\log|x| - \log|x+1|]_{2}^{3} = [\log|\frac{x}{x+1}|]_{2}^{3}$.Evaluating at the limits: $I = \log(\frac{3}{4}) - \log(\frac{2}{3}) = \log(\frac{3}{4} \div \frac{2}{3}) = \log(\frac{3}{4} 	imes \frac{3}{2}) = \log(\frac{9}{8})$.

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

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