MCQ
$\int_1^{\sqrt 3 } {\frac{1}{{1 + {x^2}}}dx} $ is equal to
  • $\pi /12$
  • B
    $\pi /6$
  • C
    $\pi /4$
  • D
    $\pi /3$

Answer

Correct option: A.
$\pi /12$
a
(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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