Answer
$$\frac{{2x - 1}}{4}\sqrt {4{x^2} - 4x - 3} - \ln \left| {2x - 1 + \sqrt {{{\left( {2x - 1} \right)}^2} - 4} } \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\sqrt {4{x^2} - 4x - 3} dx} \cr
& {\text{Completing the square}} \cr
& = \int {\sqrt {\left( {4{x^2} - 4x + 1} \right) - 41} dx} \cr
& = \int {\sqrt {{{\left( {2x - 1} \right)}^2} - 4} dx} \cr
& {\text{Rewrite the integrand}} \cr
& = \frac{1}{2}\int {\sqrt {{{\left( {2x - 1} \right)}^2} - 4} \left( 2 \right)dx} \cr
& {\text{Let }}u = 2x - 1,{\text{ }}du = dx \cr
& = \frac{1}{2}\int {\sqrt {{u^2} - 1} du} \cr
& {\text{Use the Table of Integrals on the Reference Pages }}\left( {{\text{510}}} \right) \cr
& *\int {\sqrt {{u^2} - {a^2}} du} = \frac{u}{2}\sqrt {{u^2} + {a^2}} - \frac{{{a^2}}}{2}\ln \left| {u + \sqrt {{u^2} - {a^2}} } \right| + C,{\text{ so}} \cr
& \frac{1}{2}\int {\sqrt {{u^2} - 1} du} = \frac{u}{4}\sqrt {{u^2} - 4} - \frac{4}{4}\ln \left| {u + \sqrt {{u^2} - 4} } \right| + C \cr
& {\text{Write in terms of }}x \cr
& = \frac{{2x - 1}}{4}\sqrt {{{\left( {2x - 1} \right)}^2} - 4} - \ln \left| {2x - 1 + \sqrt {{{\left( {2x - 1} \right)}^2} - 4} } \right| + C \cr
& = \frac{{2x - 1}}{4}\sqrt {4{x^2} - 4x - 3} - \ln \left| {2x - 1 + \sqrt {{{\left( {2x - 1} \right)}^2} - 4} } \right| + C \cr} $$
