A) \[\pi /12\]
B) \[\pi /6\]
C) \[\pi /4\]
D) \[\pi /3\]
Correct Answer: A
Solution :
\[\int_{1}^{\sqrt{3}}{\frac{1}{1+{{x}^{2}}}dx=[{{\tan }^{-1}}x]_{\,1}^{\sqrt{3}}=\frac{\pi }{3}-\frac{\pi }{4}=\frac{\pi }{12}}\].You need to login to perform this action.
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