Answer
$$\sqrt {{x^2} - 1} \ln x - \sqrt {{x^2} - 1} + {\sec ^{ - 1}}x + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{x\ln x}}{{\sqrt {{x^2} - 1} }}} dx \cr
& {\text{Integrate by parts, }} \cr
& {\text{Let }}u = \ln x,{\text{ }}du = \frac{1}{x}dx,{\text{ }}dv = \frac{x}{{\sqrt {{x^2} - 1} }}dx,{\text{ }}v = \sqrt {{x^2} - 1} \cr
& {\text{Using }}\int {udv} = uv - \int {vdu} \cr
& \int {\frac{{x\ln x}}{{\sqrt {{x^2} - 1} }}} dx = \sqrt {{x^2} - 1} \ln x - \int {\sqrt {{x^2} - 1} } \left( {\frac{1}{x}} \right)dx \cr
& \int {\frac{{x\ln x}}{{\sqrt {{x^2} - 1} }}} dx = \sqrt {{x^2} - 1} \ln x - \int {\frac{{\sqrt {{x^2} - 1} }}{x}} dx{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{* Integrating }}\int {\frac{{\sqrt {{x^2} - 1} }}{x}} dx,{\text{ let }}x = \sec \theta ,{\text{ }}dx = \sec \theta \tan \theta d\theta \cr
& \int {\frac{{\sqrt {{x^2} - 1} }}{x}} dx = \int {\frac{{\sqrt {{{\sec }^2}\theta - 1} }}{{\sec \theta }}\left( {\sec \theta \tan \theta } \right)} d\theta \cr
& = \int {\sqrt {{{\tan }^2}\theta } \tan \theta } d\theta \cr
& = \int {{{\tan }^2}\theta } d\theta \cr
& = \int {\left( {{{\sec }^2}\theta - 1} \right)} d\theta \cr
& = \tan \theta - \theta + C \cr
& {\text{Write in terms of }}x \cr
& = \sqrt {{x^2} - 1} - {\sec ^{ - 1}}x + C \cr
& {\text{Substituting the previous result into }}\left( {\bf{1}} \right) \cr
& \int {\frac{{x\ln x}}{{\sqrt {{x^2} - 1} }}} dx = \sqrt {{x^2} - 1} \ln x - \sqrt {{x^2} - 1} + {\sec ^{ - 1}}x + C \cr} $$
