Answer
$$\frac{1}{4}{\tan ^{ - 1}}\left( {4x + 2} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{16{x^2} + 16x + 5}}} \cr
& {\text{Completing the square for }}16{x^2} + 16x + 5 \cr
& 16{x^2} + 16x + 5 = \left( {16{x^2} + 16x + 4} \right) + 1 \cr
& 16{x^2} + 16x + 5 = {\left( {4x + 2} \right)^2} + 1 \cr
& \int {\frac{{dx}}{{16{x^2} + 16x + 5}}} = \int {\frac{{dx}}{{{{\left( {4x + 2} \right)}^2} + 1}}} \cr
& \cr
& {\text{Write in terms of }}u \cr
& {\text{Substitute }}u = 4x + 2,{\text{ }}\,\,\,du = 4dx \cr
& \int {\frac{{dx}}{{{{\left( {4x + 2} \right)}^2} + 1}}} = \int {\frac{{\left( {1/4} \right)du}}{{{u^2} + 1}}} \cr
& = \frac{1}{4}\int {\frac{{du}}{{{u^2} + 1}}} \cr
& {\text{Integrate}} \cr
& = \frac{1}{4}{\tan ^{ - 1}}u + C \cr
& \cr
& {\text{Write in terms of }}x \cr
& = \frac{1}{4}{\tan ^{ - 1}}\left( {4x + 2} \right) + C \cr} $$
