Answer
$$A = \pi $$
Work Step by Step
$$\eqalign{
& y = \frac{1}{{{x^2} + 1}},{\text{ }} - \infty < x < \infty \cr
& {\text{The area is given by}} \cr
& A = \int_{ - \infty }^\infty {\frac{1}{{{x^2} + 1}}} dx \cr
& {\text{By symmetry}} \cr
& A = 2\int_0^\infty {\frac{1}{{{x^2} + 1}}} dx \cr
& {\text{By the Definition of Improper Integrals with Infinite }} \cr
& {\text{Integration Limits}} \cr
& A = 2\mathop {\lim }\limits_{b \to \infty } \int_0^b {\frac{1}{{{x^2} + 1}}} dx \cr
& {\text{Integrate}} \cr
& A = 2\mathop {\lim }\limits_{b \to \infty } \left[ {\arctan x} \right]_0^b \cr
& A = 2\mathop {\lim }\limits_{b \to \infty } \left[ {\arctan b - \arctan 0} \right] \cr
& A = 2\mathop {\lim }\limits_{b \to \infty } \left[ {\arctan b} \right] \cr
& {\text{Evaluate the limit when }}b \to \infty \cr
& A = 2\left( {\arctan \infty } \right) \cr
& A = 2\left( {\frac{\pi }{2}} \right) \cr
& A = \pi \cr} $$
