int_1^{sqrt{3}} frac{1}{1 + x^2} dx is equal | vedclass

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(A) We know that the integral of $\frac{1}{1 + x^2}$ with respect to $x$ is $	an^{-1}(x)$.Applying the fundamental theorem of calculus for the defini

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$\pi / 12$

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(A) We know that the integral of $\frac{1}{1 + x^2}$ with respect to $x$ is $	an^{-1}(x)$.Applying the fundamental theorem of calculus for the definite integral:$\int_1^{\sqrt{3}} \frac{1}{1 + x^2} dx = [	an^{-1}(x)]_1^{\sqrt{3}}$$= 	an^{-1}(\sqrt{3}) - 	an^{-1}(1)$Since $	an^{-1}(\sqrt{3}) = \frac{\pi}{3}$ and $	an^{-1}(1) = \frac{\pi}{4}$,we have:$= \frac{\pi}{3} - \frac{\pi}{4}$$= \frac{4\pi - 3\pi}{12} = \frac{\pi}{12}$.

$\int_1^{\sqrt{3}} \frac{1}{1 + x^2} dx$ is equal to

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