$\int\limits_{ - 1}^1 {\cfrac{{dx}}{{{x^2} + 2x + 5}}}$
Integrals — Class 12 Maths Solution
Step-by-step Solution
Let $I = \int\limits_{ - 1}^1 {\cfrac{{dx}}{{{x^2} + 2x + 5}}} = \int\limits_{ - 1}^1 {\cfrac{{dx}}{{{x^2} + 2x + 1 + 4}} = \int\limits_{ - 1}^1 {\cfrac{{dx}}{{{{\left( {x + 1} \right)}^2} + {2^2}}}} }$
$= \cfrac{1}{2}\left[ {{{\tan }^{ - 1}}\cfrac{{x + 1}}{2}} \right]_{ - 1}^1 = \cfrac{1}{2}\left[ {{{\tan }^{ - 1}}\left( 1 \right) - {{\tan }^{ - 1}}\left( 0 \right)} \right] = \cfrac{1}{2} \times \cfrac{\pi }{4} = \cfrac{\pi }{8}$
***************************************
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Integrals. Curated by Sachin Sharma. Free for all students.