Answer
$$\frac{1}{4}\ln \left( {2{x^2} + 2x + 4} \right) - \frac{1}{2}{\tan ^{ - 1}}\left( {x + \frac{1}{2}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{2x + 3}}{{4{x^2} + 4x + 5}}} dx \cr
& {\text{completing the square for }}4{x^2} + 4x + 5 \cr
& 4{x^2} + 4x + 5 = 4\left( {{x^2} + x + \frac{1}{4}} \right) + 5 - 1 \cr
& 4{x^2} + 4x + 5 = 4{\left( {x + 1/2} \right)^2} + 4 \cr
& \int {\frac{{2x + 3}}{{4{x^2} + 4x + 5}}} dx = \int {\frac{{2x + 3}}{{4{{\left( {x + 1/2} \right)}^2} + 4}}} dx \cr
& \cr
& {\text{Let }}u = x + 1/2,\,\,\,\,du = dx \cr
& = \int {\frac{{2\left( {u - 1/2} \right) + 3}}{{4{u^2} + 4}}} du \cr
& = \int {\frac{{2u - 1 + 3}}{{4{u^2} + 4}}} du = \int {\frac{{2u - 2}}{{4{u^2} + 4}}} du \cr
& = \int {\frac{{u - 1}}{{2{u^2} + 2}}} du \cr
& {\text{split the numerator}} \cr
& = \frac{1}{4}\int {\frac{{4u}}{{2{u^2} + 2}}} du - \frac{1}{2}\int {\frac{1}{{{u^2} + 1}}} du \cr
& {\text{Integrate}} \cr
& = \frac{1}{4}\ln \left( {2{u^2} + 2} \right) - \frac{1}{2}{\tan ^{ - 1}}u + C \cr
& \cr
& {\text{Write in terms of }}x,{\text{ replace }}x + 1/2{\text{ for }}u \cr
& = \frac{1}{4}\ln \left( {2{{\left( {x + 1/2} \right)}^2} + 2} \right) - \frac{1}{2}{\tan ^{ - 1}}\left( {x + \frac{1}{2}} \right) + C \cr
& = \frac{1}{4}\ln \left( {2{x^2} + 2x + 4} \right) - \frac{1}{2}{\tan ^{ - 1}}\left( {x + \frac{1}{2}} \right) + C \cr} $$
