Answer
$$\sqrt {{x^2} - 1} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{xdx}}{{\sqrt {{x^2} - 1} }}} \cr
& {\text{Integrate by substitution method}} \cr
& {\text{Let }}u = {x^2} - 1,\,\,\,\,du = 2xdx,\,\,\,\,dx = \frac{{du}}{{2x}} \cr
& {\text{Then}}{\text{,}} \cr
& \int {\frac{{xdx}}{{\sqrt {{x^2} - 1} }}} = \int {\frac{x}{{\sqrt u }}\left( {\frac{{du}}{{2x}}} \right)} \cr
& {\text{Cancel common factor }}x \cr
& = \frac{1}{2}\int {\frac{1}{{\sqrt u }}du} \cr
& = \frac{1}{2}\int {\frac{1}{{{u^{1/2}}}}du} \cr
& = \frac{1}{2}\int {{u^{ - 1/2}}du} \cr
& {\text{integrating}} \cr
& = \frac{1}{2}\left( {\frac{{{u^{1/2}}}}{{1/2}}} \right) + C \cr
& = \sqrt u + C \cr
& {\text{write in terms of }}x;{\text{ replace }}u = {x^2} - 1 \cr
& = \sqrt {{x^2} - 1} + C \cr} $$
