Answer
$$A = 4\pi $$
Work Step by Step
$$\eqalign{
& y = \frac{8}{{{x^2} + 4}},{\text{ }} - \infty < x < \infty \cr
& {\text{The area is given by}} \cr
& A = \int_{ - \infty }^\infty {\frac{8}{{{x^2} + 4}}} dx \cr
& {\text{By symmetry}} \cr
& A = 2\int_0^\infty {\frac{8}{{{x^2} + 4}}} dx \cr
& {\text{By the Definition of Improper Integrals with Infinite }} \cr
& {\text{Integration Limits}} \cr
& A = 2\mathop {\lim }\limits_{b \to \infty } \int_0^b {\frac{8}{{{x^2} + 4}}} dx \cr
& {\text{Integrate}} \cr
& A = 16\mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{2}\arctan \left( {\frac{x}{2}} \right)} \right]_0^b \cr
& A = 8\mathop {\lim }\limits_{b \to \infty } \left[ {\arctan \left( {\frac{b}{2}} \right) - \arctan \left( {\frac{0}{2}} \right)} \right] \cr
& A = 8\mathop {\lim }\limits_{b \to \infty } \left[ {\arctan \left( {\frac{b}{2}} \right)} \right] \cr
& {\text{Evaluate the limit when }}b \to \infty \cr
& A = 8\left( {\arctan \left( {\frac{\infty }{2}} \right)} \right) \cr
& A = 8\left( {\frac{\pi }{2}} \right) \cr
& A = 4\pi \cr} $$
