Question
$\int\limits_1^{\sqrt 3 } {\cfrac{{dx}}{{1 + {x^2}}}}$ equals
- (a) $\cfrac{\pi }{3}$
- (b) $\cfrac{{2\pi }}{3}$
- (c) $\cfrac{\pi }{6}$
- (d) $\cfrac{\pi }{{12}}$
Integrals — Class 12 Maths Solution
Step-by-step Solution
Option d is correct
Let$I = \int\limits_1^{\sqrt 3 } {\cfrac{{dx}}{{1 + {x^2}}}} = \left[ {{{\tan }^{ - 1}}x} \right]_1^{\sqrt 3 }$
$= {\tan ^{ - 1}}\sqrt 3 - {\tan ^{ - 1}}\left( 1 \right) = \cfrac{\pi }{3} - \cfrac{\pi }{4} = \cfrac{\pi }{{12}}$
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Integrals. Curated by Sachin Sharma. Free for all students.