Answer
$$\frac{1}{2}\ln \left( {{x^2} + 4} \right) + 2{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{x + 4}}{{{x^2} + 4}}} dx \cr
& = \int {\frac{x}{{{x^2} + 4}}} dx + \int {\frac{4}{{{x^2} + 4}}} dx \cr
& {\text{write the integrands as}} \cr
& = \frac{1}{2}\int {\frac{{2x}}{{{x^2} + 4}}} dx + 4\int {\frac{1}{{{x^2} + \left( 2 \right)}}} dx \cr
& {\text{integrate by using the formulas}} \cr
& \int {\frac{{du}}{u}} = \ln \left| u \right| + C{\text{ and }}\int {\frac{1}{{{x^2} + {a^2}}}} = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{x}{a}} \right) + C \cr
& {\text{then}} \cr
& = \frac{1}{2}\ln \left| {{x^2} + 4} \right| + 4\left( {\frac{1}{2}{{\tan }^{ - 1}}\left( {\frac{x}{2}} \right)} \right) + C \cr
& {\text{simplifying}} \cr
& = \frac{1}{2}\ln \left( {{x^2} + 4} \right) + 2{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + C \cr} $$
