Evaluate the following integral: int_{1}^{2} | vedclass

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(B) Let $I = \int_{1}^{2} \frac{x \, dx}{(x+1)(x+2)}$.Using partial fractions,we decompose the integrand:$\frac{x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac

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$\log \left( \frac{32}{27} \right)$

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(B) Let $I = \int_{1}^{2} \frac{x \, dx}{(x+1)(x+2)}$.Using partial fractions,we decompose the integrand:$\frac{x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$$x = A(x+2) + B(x+1)$Setting $x = -1$,we get $A = -1$. Setting $x = -2$,we get $B = 2$.So,$\frac{x}{(x+1)(x+2)} = \frac{-1}{x+1} + \frac{2}{x+2}$.The indefinite integral is:$\int \left( \frac{-1}{x+1} + \frac{2}{x+2} ight) dx = -\log |x+1| + 2 \log |x+2| = F(x)$.Applying the Fundamental Theorem of Calculus:$I = F(2) - F(1)$$I = [-\log(3) + 2 \log(4)] - [-\log(2) + 2 \log(3)]$$I = -\log(3) + 2 \log(4) + \log(2) - 2 \log(3)$$I = 2 \log(4) + \log(2) - 3 \log(3)$$I = \log(16) + \log(2) - \log(27)$$I = \log \left( \frac{16 	imes 2}{27} ight) = \log \left( \frac{32}{27} ight)$.

Evaluate the following integral: $\int_{1}^{2} \frac{x \, dx}{(x+1)(x+2)}$

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