Answer
$$A = \frac{\pi }{4}$$
Work Step by Step
$$\eqalign{
& y = \frac{2}{{{x^2} + 4x + 8}} \cr
& {\text{The area of the region is given by}} \cr
& A = \int_{ - 2}^0 {\frac{2}{{{x^2} + 4x + 8}}} dx \cr
& {\text{Completing the square}} \cr
& A = \int_{ - 2}^0 {\frac{2}{{{x^2} + 4x + 4 + 4}}} dx \cr
& A = 2\int_{ - 2}^0 {\frac{1}{{{{\left( {x + 2} \right)}^2} + 4}}} dx \cr
& {\text{Let }}u = x + 2,{\text{ }}du = dx \cr
& {\text{The new limits of integration are}} \cr
& x = 0,{\text{ }}u = 0 + 2 = 2 \cr
& x = - 2,{\text{ }}u = - 2 + 2 = 0 \cr
& {\text{Substituting}} \cr
& A = 2\int_{ - 2}^0 {\frac{1}{{{{\left( {x + 2} \right)}^2} + 4}}} dx = 2\int_0^2 {\frac{1}{{{u^2} + 4}}} dx \cr
& {\text{Integrate using basic integration rules}} \cr
& A = 2\left[ {\frac{1}{2}{{\tan }^{ - 1}}\frac{u}{2}} \right]_0^2 \cr
& A = {\tan ^{ - 1}}\frac{2}{2} - {\tan ^{ - 1}}\frac{0}{2} \cr
& {\text{Simplify}} \cr
& A = \frac{\pi }{4} - 0 \cr
& A = \frac{\pi }{4} \cr} $$
