#### Answer

$$\frac{1}{2}\ln \left( {{x^2} + 8x + 25} \right) + C$$

#### Work Step by Step

$$\eqalign{
& \int {\frac{{x + 4}}{{{x^2} + 8x + 25}}} dx \cr
& {\text{completing the square}} \cr
& = \int {\frac{{x + 4}}{{{x^2} + 8x + 16 + 9}}} dx \cr
& = \int {\frac{{x + 4}}{{{{\left( {x + 4} \right)}^2} + 9}}} dx \cr
& {\text{substitute }}u = x + 4,{\text{ }}du = dx \cr
& = \int {\frac{u}{{{u^2} + 9}}} du \cr
& = \frac{1}{2}\int {\frac{{2u}}{{{u^2} + 9}}} du \cr
& {\text{find the antiderivative}} \cr
& = \frac{1}{2}\ln \left| {{u^2} + 9} \right| + C \cr
& = \frac{1}{2}\ln \left( {{u^2} + 9} \right) + C \cr
& {\text{replace back }}u = x + 4 \cr
& = \frac{1}{2}\ln \left( {{{\left( {x + 4} \right)}^2} + 9} \right) + C \cr
& = \frac{1}{2}\ln \left( {{x^2} + 8x + 25} \right) + C \cr} $$