Answer
$$ - \frac{1}{2}\ln \frac{{2 + \sqrt {{x^2} + 4x + 8} }}{{\left| {x - 2} \right|}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{\left( {x + 2} \right)\sqrt {{x^2} + 4x + 8} }}} \cr
& {\text{Completing the square}} \cr
& = \int {\frac{{dx}}{{\left( {x + 2} \right)\sqrt {\left( {{x^2} + 4x + 4} \right) + 4} }}} \cr
& = \int {\frac{{dx}}{{\left( {x + 2} \right)\sqrt {{{\left( {x + 2} \right)}^2} + 4} }}} \cr
& {\text{Let }}u = x - 2,{\text{ }}du = dx \cr
& = \int {\frac{{du}}{{u\sqrt {{u^2} + 4} }}} \cr
& {\text{Use Theorem 5}}{\text{.20 }}\int {\frac{{du}}{{u\sqrt {{u^2} + {a^2}} }} = } - \frac{1}{a}\ln \frac{{a + \sqrt {{u^2} + {a^2}} }}{{\left| u \right|}} + C \cr
& \int {\frac{{du}}{{u\sqrt {{u^2} + 4} }}} = - \frac{1}{2}\ln \frac{{2 + \sqrt {{u^2} + 4} }}{{\left| u \right|}} + C \cr
& {\text{Write in terms of }}x \cr
& = - \frac{1}{2}\ln \frac{{2 + \sqrt {{{\left( {x - 2} \right)}^2} + 4} }}{{\left| {x - 2} \right|}} + C \cr
& = - \frac{1}{2}\ln \frac{{2 + \sqrt {{x^2} + 4x + 8} }}{{\left| {x - 2} \right|}} + C \cr} $$
