Evaluate the definite integral int_{1}^{sqrt{ | vedclass

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Question

(A) We know that the antiderivative of $\frac{1}{1+x^2}$ is $	an^{-1} x$.By the second fundamental theorem of calculus,we have:$\int_{1}^{\sqrt{3}} \

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$\frac{\pi}{12}$

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(A) We know that the antiderivative of $\frac{1}{1+x^2}$ is $	an^{-1} x$.By the second fundamental theorem of calculus,we have:$\int_{1}^{\sqrt{3}} \frac{d x}{1+x^{2}} = [	an^{-1} x]_{1}^{\sqrt{3}}$Substitute the upper and lower limits:$= 	an^{-1}(\sqrt{3}) - 	an^{-1}(1)$We know that $	an^{-1}(\sqrt{3}) = \frac{\pi}{3}$ and $	an^{-1}(1) = \frac{\pi}{4}$.$= \frac{\pi}{3} - \frac{\pi}{4}$$= \frac{4\pi - 3\pi}{12} = \frac{\pi}{12}$Thus,the correct option is $A$.

Evaluate the definite integral $\int_{1}^{\sqrt{3}} \frac{d x}{1+x^{2}}$.

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