[DCE 2002]
A) $ \pi /12 $
B) $ \pi /6 $
C) $ \pi /4 $
D) $ \pi /3 $
Correct Answer: A
$ \int_1^{\sqrt{3}}{\frac{1}{1+x^{2}}dx=[{{\tan }^{-1}}x] _{1}^{\sqrt{3}}=\frac{\pi }{3}-\frac{\pi }{4}=\frac{\pi }{12}} $ .
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