int_{2}^{3} frac{(x+2)^2}{2x^2-10x+53} dx = | vedclass

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(C) Let $I = \int_{2}^{3} \frac{(x+2)^2}{2x^2-10x+53} dx$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we have $a+b = 2+3 = 5

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(C) Let $I = \int_{2}^{3} \frac{(x+2)^2}{2x^2-10x+53} dx$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we have $a+b = 2+3 = 5$. So,$I = \int_{2}^{3} \frac{(5-x+2)^2}{2(5-x)^2-10(5-x)+53} dx = \int_{2}^{3} \frac{(7-x)^2}{2(25-10x+x^2)-50+10x+53} dx$. Simplifying the denominator: $2x^2-20x+50-50+10x+53 = 2x^2-10x+53$. Thus,$I = \int_{2}^{3} \frac{(7-x)^2}{2x^2-10x+53} dx$. Adding the two expressions for $I$: $2I = \int_{2}^{3} \frac{(x+2)^2+(7-x)^2}{2x^2-10x+53} dx = \int_{2}^{3} \frac{x^2+4x+4+49-14x+x^2}{2x^2-10x+53} dx = \int_{2}^{3} \frac{2x^2-10x+53}{2x^2-10x+53} dx = \int_{2}^{3} 1 dx = [x]_{2}^{3} = 3-2 = 1$. Therefore,$I = \frac{1}{2}$.

$\int_{2}^{3} \frac{(x+2)^2}{2x^2-10x+53} dx = $

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