The value of the integral int_1^2 frac{x , dx | vedclass

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(D) To evaluate the integral $\int_1^2 \frac{x}{(x+2)(x+3)} \, dx$,we use partial fractions. Let $\frac{x}{(x+2)(x+3)} = \frac{A}{x+2} + \frac{B}{x+3}

Vedclass · Accepted Answer

$\log \left(\frac{1125}{1024}\right)$

Vedclass · Answer

(D) To evaluate the integral $\int_1^2 \frac{x}{(x+2)(x+3)} \, dx$,we use partial fractions. Let $\frac{x}{(x+2)(x+3)} = \frac{A}{x+2} + \frac{B}{x+3}$. Multiplying by $(x+2)(x+3)$,we get $x = A(x+3) + B(x+2)$. Setting $x = -2$,we get $-2 = A(1)$,so $A = -2$. Setting $x = -3$,we get $-3 = B(-1)$,so $B = 3$. Thus,$\int_1^2 \left( \frac{-2}{x+2} + \frac{3}{x+3} \right) \, dx = [-2 \log|x+2| + 3 \log|x+3|]_1^2$. Evaluating at the limits: $(-2 \log 4 + 3 \log 5) - (-2 \log 3 + 3 \log 4) = -5 \log 4 + 3 \log 5 + 2 \log 3 = \log(5^3 \cdot 3^2 / 4^5) = \log(125 \cdot 9 / 1024) = \log(\frac{1125}{1024})$. Therefore,the correct option is $D$.

The value of the integral $\int_1^2 \frac{x \, dx}{(x+2)(x+3)}$ is

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