Answer
$$\frac{1}{4}\ln \left| {{x^2} + 3x - \frac{7}{4}} \right| - \frac{1}{8}\ln \left| {\frac{{2x + 1}}{{2x + 5}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{2x + 1}}{{4{x^2} + 12x - 7}}} dx \cr
& {\text{Factor the denominator}} \cr
& = \int {\frac{{2x + 1}}{{4\left( {{x^2} + 3x - 7/4} \right)}}} dx \cr
& {\text{Complete the square}} \cr
& = \frac{1}{4}\int {\frac{{2x + 1}}{{{x^2} + 3x + 9/4 - 9/4 - 7/4}}} dx \cr
& = \frac{1}{4}\int {\frac{{2x + 1}}{{{{\left( {x + 3/2} \right)}^2} - 4}}} dx \cr
& {\text{Let }}u = x + \frac{3}{2} \to x = u - \frac{3}{2},{\text{ }}dx = du \cr
& = \frac{1}{4}\int {\frac{{2\left( {u - 3/2} \right) + 1}}{{{u^2} - 4}}} du \cr
& = \frac{1}{4}\int {\frac{{2u - 3 + 1}}{{{u^2} - 4}}} du \cr
& = \frac{1}{2}\int {\frac{{u - 1}}{{{u^2} - 4}}} du \cr
& = \frac{1}{2}\int {\frac{u}{{{u^2} - 4}}} du - \frac{1}{2}\int {\frac{1}{{{u^2} - 4}}} du \cr
& {\text{Using the formula 6 }}\int {\frac{{dx}}{{{x^2} - {a^2}}}} = \frac{1}{{2a}}\ln \left| {\frac{{x - a}}{{x + a}}} \right| + C \cr
& = \frac{1}{4}\ln \left| {{u^2} - 4} \right| - \frac{1}{2}\left( {\frac{1}{{2\left( 2 \right)}}\ln \left| {\frac{{u - 2}}{{u + 2}}} \right|} \right) + C \cr
& = \frac{1}{4}\ln \left| {{u^2} - 4} \right| - \frac{1}{8}\ln \left| {\frac{{u - 2}}{{u + 2}}} \right| + C \cr
& {\text{Write in terms of }}x,u = x + \frac{3}{2} \cr
& = \frac{1}{4}\ln \left| {{{\left( {x + \frac{3}{2}} \right)}^2} - 4} \right| - \frac{1}{8}\ln \left| {\frac{{x + \frac{3}{2} - 2}}{{x + \frac{3}{2} + 2}}} \right| + C \cr
& = \frac{1}{4}\ln \left| {{x^2} + 3x + \frac{9}{4} - 4} \right| - \frac{1}{8}\ln \left| {\frac{{2x + 1}}{{2x + 5}}} \right| + C \cr
& = \frac{1}{4}\ln \left| {{x^2} + 3x - \frac{7}{4}} \right| - \frac{1}{8}\ln \left| {\frac{{2x + 1}}{{2x + 5}}} \right| + C \cr} $$
