Answer
$$A = \frac{\pi }{2}$$
Work Step by Step
$$\eqalign{
& y = \frac{1}{{{x^2} + 1}},{\text{ }}y = 0,{\text{ }}x = 1 \cr
& {\text{From the graph we can note that }}\frac{1}{{{x^2} + 1}} > 0{\text{ on }}\left[ { - 1,1} \right] \cr
& {\text{Then, the area is given by:}} \cr
& A = \int_{ - 1}^1 {\left( {\frac{1}{{{x^2} + 1}} - 0} \right)} dx \cr
& A = \int_{ - 1}^1 {\frac{1}{{{x^2} + 1}}} dx \cr
& {\text{By symmetry}} \cr
& A = 2\int_0^1 {\frac{1}{{{x^2} + 1}}} dx \cr
& {\text{Integrate and evaluate}} \cr
& A = 2\left[ {{{\tan }^{ - 1}}x} \right]_0^1 \cr
& A = 2\left[ {{{\tan }^{ - 1}}\left( 1 \right)} \right] \cr
& A = 2\left( {\frac{\pi }{4}} \right) \cr
& A = \frac{\pi }{2} \cr} $$
