Answer
$$\frac{1}{2}{\sec ^{ - 1}}\left| {2x + 1} \right| + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{\left( {2x + 1} \right)\sqrt {4x + 4{x^2}} }}} \cr
& {\text{complete the square for }}4x + 4{x^2} \cr
& 4x + 4{x^2} = 4{x^2} + 4x + 1 - 1 \cr
& 4x + 4{x^2} = {\left( {2x + 1} \right)^2} - 1 \cr
& {\text{then}}{\text{,}} \cr
& \int {\frac{{dx}}{{\left( {2x + 1} \right)\sqrt {4x + 4{x^2}} }}} = \int {\frac{{dx}}{{\left( {2x + 1} \right)\sqrt {{{\left( {2x + 1} \right)}^2} - 1} }}} \cr
& {\text{integrate by the substitution method}} \cr
& {\text{set }}u = 2x + 1{\text{ then }}du = 2dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{{dx}}{{\left( {2x + 1} \right)\sqrt {{{\left( {2x + 1} \right)}^2} - 1} }}} = \int {\frac{{du/2}}{{u\sqrt {{u^2} - 1} }}} \cr
& {\text{integrate by using the formula }}\int {\frac{{du}}{{u\sqrt {{u^2} - {a^2}} }} = \frac{1}{a}{{\sec }^{ - 1}}\left| {\frac{u}{a}} \right| + C},.{\text{ }}a = 1 \cr
& \frac{1}{2}\int {\frac{{du}}{{u\sqrt {{u^2} - {7^2}} }}} = \frac{1}{2}\left( {\frac{1}{1}{{\sec }^{ - 1}}\left| {\frac{u}{1}} \right|} \right) + C \cr
& = \frac{1}{2}{\sec ^{ - 1}}\left| u \right| + C \cr
& {\text{replace }}2x + 1{\text{ for }}u \cr
& = \frac{1}{2}{\sec ^{ - 1}}\left| {2x + 1} \right| + C \cr} $$
