The value of int_{0}^{1} frac{2x^2 + 3x + 3}{ | vedclass

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(D) Let $I = \int_{0}^{1} \frac{2x^2 + 3x + 3}{(x + 1)(x^2 + 2x + 2)} dx$. Using partial fractions,we write $\frac{2x^2 + 3x + 3}{(x + 1)(x^2 + 2x + 2

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(D) Let $I = \int_{0}^{1} \frac{2x^2 + 3x + 3}{(x + 1)(x^2 + 2x + 2)} dx$. Using partial fractions,we write $\frac{2x^2 + 3x + 3}{(x + 1)(x^2 + 2x + 2)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+2x+2}$. Equating coefficients,we get $A(x^2+2x+2) + (Bx+C)(x+1) = 2x^2+3x+3$. For $x = -1$,$A(1-2+2) = 2-3+3 \implies A = 2$. Comparing coefficients of $x^2$,$A+B = 2 \implies B = 0$. Comparing constants,$2A+C = 3 \implies 4+C = 3 \implies C = -1$. So,$I = \int_{0}^{1} (\frac{2}{x+1} - \frac{1}{x^2+2x+2}) dx = \int_{0}^{1} (\frac{2}{x+1} - \frac{1}{(x+1)^2+1}) dx$. $I = [2 \ln|x+1| - 	an^{-1}(x+1)]_{0}^{1} = (2 \ln 2 - 	an^{-1} 2) - (2 \ln 1 - 	an^{-1} 1) = 2 \ln 2 - 	an^{-1} 2 + \frac{\pi}{4}$. Since $	an^{-1} 2 + \cot^{-1} 2 = \frac{\pi}{2}$,we have $\frac{\pi}{4} - 	an^{-1} 2 = \frac{\pi}{4} - (\frac{\pi}{2} - \cot^{-1} 2) = \cot^{-1} 2 - \frac{\pi}{4}$. Thus,$I = 2 \ln 2 + \cot^{-1} 2 - \frac{\pi}{4}$. Both options $A$ and $B$ are equivalent. Therefore,the correct answer is $D$.

The value of $\int_{0}^{1} \frac{2x^2 + 3x + 3}{(x + 1)(x^2 + 2x + 2)} dx$ is:

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