Answer
$$A = 2\ln 3$$
Work Step by Step
$$\eqalign{
& {\text{The area if the region is given by}} \cr
& A = \int_4^{ + \infty } {\frac{8}{{{x^2} - 4}}} dx \cr
& {\text{Using the definition of improper integrals}} \cr
& A = \mathop {\lim }\limits_{b \to + \infty } \int_4^{ + \infty } {\frac{8}{{{x^2} - 4}}} dx \cr
& A = 8\mathop {\lim }\limits_{b \to + \infty } \left[ {\frac{1}{{{x^2} - 4}}} \right]_4^b \cr
& {\text{Integrate using }}\int {\frac{{dx}}{{{x^2} - {a^2}}} = \frac{1}{{2a}}\ln \left| {\frac{{x - a}}{{x + a}}} \right| + C} \cr
& A = 8\mathop {\lim }\limits_{b \to + \infty } \left[ {\frac{1}{{2\left( 2 \right)}}\ln \left| {\frac{{x - 2}}{{x + 2}}} \right|} \right]_4^b \cr
& A = 2\mathop {\lim }\limits_{b \to + \infty } \left[ {\ln \left| {\frac{{x - 2}}{{x + 2}}} \right|} \right]_4^b \cr
& A = 2\mathop {\lim }\limits_{b \to + \infty } \left[ {\ln \left| {\frac{{x - 2}}{{x + 2}}} \right|} \right]_4^b \cr
& A = 2\mathop {\lim }\limits_{b \to + \infty } \left[ {\ln \left| {\frac{{b - 2}}{{b + 2}}} \right| - \ln \left| {\frac{{4 - 2}}{{4 + 2}}} \right|} \right] \cr
& {\text{Evaluate the limit}} \cr
& A = 2\left[ {\ln \left| {\frac{{\infty - 2}}{{\infty + 2}}} \right| - \ln \left| {\frac{1}{3}} \right|} \right] \cr
& A = 2\left[ {\ln \left| { - 1} \right| - \ln \left| {\frac{1}{3}} \right|} \right] \cr
& A = 2\left[ { - \ln \left| {\frac{1}{3}} \right|} \right] \cr
& A = 2\ln 3 \cr} $$
