Answer
$${\tan ^{ - 1}}2x - \frac{1}{{4{x^2} + 1}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{8{x^2} + 8x + 2}}{{{{\left( {4{x^2} + 1} \right)}^2}}}} dx \cr
& {\text{Decomposing the integrand }}\frac{{8{x^2} + 8x + 2}}{{{{\left( {4{x^2} + 1} \right)}^2}}}{\text{ into partial fractions}} \cr
& \frac{{8{x^2} + 8x + 2}}{{{{\left( {4{x^2} + 1} \right)}^2}}} = \frac{{Ax + B}}{{4{x^2} + 1}} + \frac{{Cx + D}}{{{{\left( {4{x^2} + 1} \right)}^2}}} \cr
& {\text{multiply by }}{\left( {4{x^2} + 1} \right)^2}{\text{ and simplify}} \cr
& 8{x^2} + 8x + 2 = \left( {Ax + B} \right)\left( {4{x^2} + 1} \right) + Cx + D \cr
& {\text{Expanding the equation }} \cr
& 8{x^2} + 8x + 2 = 4A{x^3} + Ax + 4B{x^2} + B + Cx + D \cr
& {\text{Group terms}} \cr
& 8{x^2} + 8x + 2 = 4A{x^3} + 4B{x^2} + Ax + Cx + B + D \cr
& {\text{Equating coefficients}} \cr
& A = 0,\,\,\,\,4B = 8 \to B = 2\,\,,\,\,\,\,A + C = 8 \to C = 8,\,\,\,\,B + D = 2 \to D = 0 \cr
& \cr
& \frac{{8{x^2} + 8x + 2}}{{{{\left( {4{x^2} + 1} \right)}^2}}} = \frac{{Ax + B}}{{4{x^2} + 1}} + \frac{{Cx + D}}{{{{\left( {4{x^2} + 1} \right)}^2}}} = \frac{2}{{4{x^2} + 1}} + \frac{{8x}}{{{{\left( {4{x^2} + 1} \right)}^2}}} \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& \int {\frac{{8{x^2} + 8x + 2}}{{{{\left( {4{x^2} + 1} \right)}^2}}}} dx = \int {\left( {\frac{2}{{4{x^2} + 1}} + \frac{{8x}}{{{{\left( {4{x^2} + 1} \right)}^2}}}} \right)} dx \cr
& = \int {\frac{2}{{4{x^2} + 1}}} dx + \int {{{\left( {4{x^2} + 1} \right)}^{ - 2}}\left( {8x} \right)} dx \cr
& {\text{integrating}} \cr
& = 2\left( {\frac{1}{2}{{\tan }^{ - 1}}2x} \right) + \frac{{{{\left( {4{x^2} + 1} \right)}^{ - 1}}}}{{ - 1}} + C \cr
& = {\tan ^{ - 1}}2x - \frac{1}{{4{x^2} + 1}} + C \cr} $$
