$\int\limits_1^{\sqrt 3 } {\cfrac{{dx}}{{1 + {x^2}}}}$ equals

Option d is correct

Let$I = \int\limits_1^{\sqrt 3 } {\cfrac{{dx}}{{1 + {x^2}}}} = \left[ {{{\tan }^{ - 1}}x} \right]_1^{\sqrt 3 }$

$= {\tan ^{ - 1}}\sqrt 3 - {\tan ^{ - 1}}\left( 1 \right) = \cfrac{\pi }{3} - \cfrac{\pi }{4} = \cfrac{\pi }{{12}}$