Answer
$$2\operatorname{arcsec} \left| {x - 1} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{2}{{\left( {x - 1} \right)\sqrt {{x^2} - 2x} }}} dx \cr
& {\text{Completing the square for }}{x^2} - 2x \cr
& {x^2} - 2x = {x^2} - 2x + 1 - 1 \cr
& {x^2} - 2x = {\left( {x - 1} \right)^2} - 1 \cr
& = \int {\frac{2}{{\left( {x - 1} \right)\sqrt {{x^2} - 2x + 1 - 1} }}} dx \cr
& = \int {\frac{2}{{\left( {x - 1} \right)\sqrt {{{\left( {x - 1} \right)}^2} - 1} }}} dx \cr
& {\text{Let }}u = x - 1,{\text{ }}x = u + 1,{\text{ }}dx = du,{\text{ }} \cr
& {\text{Substituting}} \cr
& \int {\frac{2}{{\left( {x - 1} \right)\sqrt {{{\left( {x - 1} \right)}^2} - 1} }}} dx = \int {\frac{2}{{u\sqrt {{u^2} - 1} }}} du \cr
& {\text{Integrate using the formula }}\int {\frac{{du}}{{u\sqrt {{u^2} - {a^2}} }} = \frac{1}{a}\operatorname{arcsec} \frac{{\left| u \right|}}{a} + C} \cr
& \left( {{\text{See the previous formula on page 378, formula 20}}} \right) \cr
& \int {\frac{2}{{u\sqrt {{u^2} - 1} }}} du = 2\left( {\frac{1}{1}} \right)\operatorname{arcsec} \frac{{\left| u \right|}}{1} + C \cr
& = 2\operatorname{arcsec} \left| u \right| + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}x - 1{\text{ for }}u \cr
& = 2\operatorname{arcsec} \left| {x - 1} \right| + C \cr} $$
