Answer
$$\frac{1}{2}{\tan ^{ - 1}}\left( {\frac{{x + 2}}{2}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{{x^2} + 4x + 8}}} dx \cr
& {\text{Completing the square}} \cr
& {\text{ = }}\int {\frac{1}{{\left( {{x^2} + 4x + 4} \right) + 4}}} dx \cr
& {\text{ = }}\int {\frac{1}{{{{\left( {x + 2} \right)}^2} + 4}}} dx \cr
& u = x + 2,{\text{ }}du = dx \cr
& \int {\frac{1}{{{{\left( {x + 2} \right)}^2} + 4}}} dx = \int {\frac{1}{{{u^2} + 4}}} du \cr
& {\text{Integrate by tables }}\int {\frac{1}{{{u^2} + {a^2}}}} du = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{u}{a}} \right) + C \cr
& \int {\frac{1}{{{u^2} + 4}}} du = \frac{1}{2}{\tan ^{ - 1}}\left( {\frac{u}{2}} \right) + C \cr
& {\text{Write in terms of }}x,{\text{ }}u = x + 2 \cr
& = \frac{1}{2}{\tan ^{ - 1}}\left( {\frac{{x + 2}}{2}} \right) + C \cr} $$
