Answer
$$\frac{1}{4}se{c^{ - 1}}\left| {\frac{{2x - 1}}{2}} \right| + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{\left( {2x - 1} \right)\sqrt {{{\left( {2x - 1} \right)}^2} - 4} }}} \cr
& {\text{use the substitution method}}{\text{.}} \cr
& u = 2x - 1,{\text{ so that }}du = 2dx \cr
& {\text{then}} \cr
& \int {\frac{{dx}}{{\left( {2x - 1} \right)\sqrt {{{\left( {2x - 1} \right)}^2} - 4} }}} = \int {\frac{{du/2}}{{u\sqrt {{u^2} - 4} }}} \cr
& = \frac{1}{2}\int {\frac{{du}}{{u\sqrt {{u^2} - 4} }}} \cr
& {\text{integrate by using the formula }}\int {\frac{{du}}{{u\sqrt {{u^2} - {a^2}} }} = \frac{1}{a}se{c^{ - 1}}\left| {\frac{u}{a}} \right| + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = 2 \cr
& = \frac{1}{2}\left( {\frac{1}{2}se{c^{ - 1}}\left| {\frac{u}{2}} \right|} \right) + C \cr
& = \frac{1}{4}se{c^{ - 1}}\left| {\frac{u}{2}} \right| + C \cr
& {\text{write in terms of }}x;{\text{ replace }}2x - 1{\text{ for }}u \cr
& = \frac{1}{4}se{c^{ - 1}}\left| {\frac{{2x - 1}}{2}} \right| + C \cr} $$
