Answer
$$\frac{1}{2}{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) - \frac{3}{{{x^2} + 4}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2} + 6x + 4}}{{{x^4} + 8{x^2} + 16}}} dx \cr
& {\text{Factor the denominator}} \cr
& = \int {\frac{{{x^2} + 6x + 4}}{{{{\left( {{x^2} + 4} \right)}^2}}}} dx \cr
& {\text{Decompose the integrand into partial fractions}} \cr
& \frac{{{x^2} + 6x + 4}}{{{{\left( {{x^2} + 4} \right)}^2}}} = \frac{{Ax + B}}{{{x^2} + 4}} + \frac{{Cx + D}}{{{{\left( {{x^2} + 4} \right)}^2}}} \cr
& {x^2} + 6x + 4 = \left( {Ax + B} \right)\left( {{x^2} + 4} \right) + Cx + D \cr
& {x^2} + 6x + 4 = A{x^3} + 4Ax + B{x^2} + 4B + Cx + D \cr
& {\text{Grouping terms}} \cr
& {x^2} + 6x + 4 = A{x^3} + B{x^2} + \left( {4Ax + Cx} \right) + \left( {4B + D} \right) \cr
& {\text{Setting a system of equations}} \cr
& A = 0 \cr
& B = 1 \cr
& 4A + C = 6 \to C = 6 \cr
& 4B + D = 4 \to D = 0 \cr
& \frac{{{x^2} + 6x + 4}}{{{{\left( {{x^2} + 4} \right)}^2}}} = \frac{{Ax + B}}{{{x^2} + 4}} + \frac{{Cx + D}}{{{{\left( {{x^2} + 4} \right)}^2}}} \cr
& \frac{{{x^2} + 6x + 4}}{{{{\left( {{x^2} + 4} \right)}^2}}} = \frac{1}{{{x^2} + 4}} + \frac{{6x}}{{{{\left( {{x^2} + 4} \right)}^2}}} \cr
& \int {\frac{{{x^2} + 6x + 4}}{{{x^4} + 8{x^2} + 16}}} dx = \int {\left( {\frac{1}{{{x^2} + 4}} + \frac{{6x}}{{{{\left( {{x^2} + 4} \right)}^2}}}} \right)} dx \cr
& = \int {\frac{1}{{{x^2} + 4}}} dx + 3\int {\frac{{2x}}{{{{\left( {{x^2} + 4} \right)}^2}}}} dx \cr
& {\text{Integrating}} \cr
& = \frac{1}{2}{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + 3\left( { - \frac{1}{{{x^2} + 4}}} \right) + C \cr
& = \frac{1}{2}{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) - \frac{3}{{{x^2} + 4}} + C \cr} $$
