int_{0}^{5} frac{d x}{x^{2}+2 x+10} = | vedclass

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(B) We have the integral $I = \int_{0}^{5} \frac{d x}{x^{2}+2 x+10}$.Completing the square in the denominator: $x^{2}+2x+10 = (x+1)^{2} + 3^{2}$.Thus,

Vedclass · Accepted Answer

$\frac{\pi}{12}$

Vedclass · Answer

(B) We have the integral $I = \int_{0}^{5} \frac{d x}{x^{2}+2 x+10}$.Completing the square in the denominator: $x^{2}+2x+10 = (x+1)^{2} + 3^{2}$.Thus,$I = \int_{0}^{5} \frac{d x}{(x+1)^{2} + 3^{2}}$.Using the formula $\int \frac{dx}{x^2+a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$,we get:$I = \left[ \frac{1}{3} 	an^{-1}(\frac{x+1}{3}) ight]_{0}^{5}$.Evaluating at the limits:$I = \frac{1}{3} [	an^{-1}(\frac{5+1}{3}) - 	an^{-1}(\frac{0+1}{3})] = \frac{1}{3} [	an^{-1}(2) - 	an^{-1}(\frac{1}{3})]$.Using the identity $	an^{-1}(x) - 	an^{-1}(y) = 	an^{-1}(\frac{x-y}{1+xy})$:$I = \frac{1}{3} 	an^{-1}(\frac{2 - 1/3}{1 + 2(1/3)}) = \frac{1}{3} 	an^{-1}(\frac{5/3}{5/3}) = \frac{1}{3} 	an^{-1}(1)$.Since $	an^{-1}(1) = \frac{\pi}{4}$,we have $I = \frac{1}{3} 	imes \frac{\pi}{4} = \frac{\pi}{12}$.

$\int_{0}^{5} \frac{d x}{x^{2}+2 x+10} = $

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