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$\int_1^{\sqrt{3}} \frac{d x}{1+x^2}$ equal
(a) $\frac{\pi}{3}$
(b) $\frac{2 \pi}{3}$
(c) $\frac{\pi}{6}$
(d) $\frac{\pi}{12}$
$\int_1^{\sqrt{3}} \frac{d x}{1+x^2}$ equal
(a) $\frac{\pi}{3}$
(b) $\frac{2 \pi}{3}$
(c) $\frac{\pi}{6}$
(d) $\frac{\pi}{12}$
Solution:
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Verified Answer
(d) $\left.\int_1^{\sqrt{3}} \frac{d x}{1+x^2}=\tan ^{-1} x\right]_1^{\sqrt{3}}=\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}$
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