Answer
$2{\tan ^{ - 1}}\sqrt {x - 1} + C$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{x\sqrt {x - 1} }}} dx \cr
& {\text{Use the substitution method}} \cr
& {\text{Let }}{u^2} = x - 1,{\text{ }}x = {u^2} + 1,{\text{ }}dx = 2udu \cr
& {\text{Substituting}} \cr
& \int {\frac{1}{{x\sqrt {x - 1} }}} dx = \int {\frac{1}{{\left( {{u^2} + 1} \right)\sqrt {{u^2}} }}} \left( {2u} \right)du \cr
& {\text{Simplify the integrand}} \cr
& = \int {\frac{{2u}}{{\left( {{u^2} + 1} \right)u}}} du \cr
& = 2\int {\frac{1}{{{u^2} + 1}}} du \cr
& {\text{Integrating}} \cr
& = 2{\tan ^{ - 1}}u + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}\sqrt {x - 1} {\text{ for }}u \cr
& = 2{\tan ^{ - 1}}\sqrt {x - 1} + C \cr} $$