Answer
$$A = \frac{{7\pi }}{{12}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{1}{{{x^2} + 1}},{\text{ }}\left[ { - 1,\sqrt 3 } \right] \cr
& \frac{1}{{{x^2} + 1}} \geqslant 0{\text{ for the interval }}\left[ { - 1,\sqrt 3 } \right],{\text{ then the area is given by}} \cr
& A = \int_{ - 1}^{\sqrt 3 } {\frac{1}{{{x^2} + 1}}} dx \cr
& {\text{Integrating}} \cr
& A = \left[ {{{\tan }^{ - 1}}x} \right]_{ - 1}^{\sqrt 3 } \cr
& {\text{Evaluating}} \cr
& A = {\tan ^{ - 1}}\left( {\sqrt 3 } \right) - {\tan ^{ - 1}}\left( { - 1} \right) \cr
& {\text{Simplifying}} \cr
& A = \frac{\pi }{3} + \frac{\pi }{4} \cr
& A = \frac{{7\pi }}{{12}} \cr} $$
