The value of int_2^3 frac{x + 1}{x^2(x - 1)} | vedclass

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(B) Let $I = \int_2^3 \frac{x + 1}{x^2(x - 1)} dx$. Using partial fractions,we write $\frac{x + 1}{x^2(x - 1)} = \frac{A}{x^2} + \frac{B}{x} + \frac{C

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$\log \frac{16}{9} - \frac{1}{6}$

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(B) Let $I = \int_2^3 \frac{x + 1}{x^2(x - 1)} dx$. Using partial fractions,we write $\frac{x + 1}{x^2(x - 1)} = \frac{A}{x^2} + \frac{B}{x} + \frac{C}{x - 1}$. Multiplying by $x^2(x - 1)$,we get $A(x - 1) + Bx(x - 1) + Cx^2 = x + 1$. Setting $x = 0$,we get $A(-1) = 1 \implies A = -1$. Setting $x = 1$,we get $C(1)^2 = 1 + 1 \implies C = 2$. Comparing the coefficients of $x^2$,we get $B + C = 0 \implies B = -2$. Thus,$I = \int_2^3 \left( -\frac{1}{x^2} - \frac{2}{x} + \frac{2}{x - 1} \right) dx$. $I = \left[ \frac{1}{x} - 2\log|x| + 2\log|x - 1| \right]_2^3$. $I = \left[ \frac{1}{x} + 2\log\left| \frac{x - 1}{x} \right| \right]_2^3$. $I = \left( \frac{1}{3} + 2\log\left( \frac{2}{3} \right) \right) - \left( \frac{1}{2} + 2\log\left( \frac{1}{2} \right) \right)$. $I = \left( \frac{1}{3} - \frac{1}{2} \right) + 2\log\left( \frac{2/3}{1/2} \right) = -\frac{1}{6} + 2\log\left( \frac{4}{3} \right)$. $I = \log\left( \frac{4}{3} \right)^2 - \frac{1}{6} = \log\frac{16}{9} - \frac{1}{6}$.

The value of $\int_2^3 \frac{x + 1}{x^2(x - 1)} dx$ is

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