int_{-1}^{0} frac{dx}{x^2 + 2x + 2} = | vedclass

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(B) Let $I = \int_{-1}^{0} \frac{dx}{x^2 + 2x + 2}$.First,complete the square in the denominator: $x^2 + 2x + 2 = (x^2 + 2x + 1) + 1 = (x + 1)^2 + 1^2

Vedclass · Accepted Answer

$\pi / 4$

Vedclass · Answer

(B) Let $I = \int_{-1}^{0} \frac{dx}{x^2 + 2x + 2}$.First,complete the square in the denominator: $x^2 + 2x + 2 = (x^2 + 2x + 1) + 1 = (x + 1)^2 + 1^2$.Thus,the integral becomes $I = \int_{-1}^{0} \frac{dx}{(x + 1)^2 + 1^2}$.Using the standard integral formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$,we get:$I = [	an^{-1}(x + 1)]_{-1}^{0}$.Evaluating the limits:$I = 	an^{-1}(0 + 1) - 	an^{-1}(-1 + 1) = 	an^{-1}(1) - 	an^{-1}(0)$.Since $	an^{-1}(1) = \frac{\pi}{4}$ and $	an^{-1}(0) = 0$,we have:$I = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.

$\int_{-1}^{0} \frac{dx}{x^2 + 2x + 2} = $

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