Calculus: Early Transcendentals (2nd Edition)

$= \frac{{{x^2}{{\sec }^{ - 1}}x}}{2} - \frac{1}{2}\sqrt {{x^2} - 1} + C$
$\begin{gathered} \int_{}^{} {x{{\sec }^{ - 1}}x\,\,\,dx} \hfill \\ \hfill \\ set\,\,\,the\,\,substitution \hfill \\ \hfill \\ u = {\sec ^{ - 1}}x\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,du = \frac{{du}}{{x\sqrt {{x^2} - 1} }} \hfill \\ dv = dx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,v = \frac{{{x^2}}}{2} \hfill \\ \hfill \\ use\,\,uv - \int_{}^{} {vdu} \hfill \\ \hfill \\ {\text{replacing the values }}{\text{in the equation}} \hfill \\ \hfill \\ = \frac{{{x^2}{{\sec }^{ - 1}}x}}{2} - \int_{}^{} {\,\left( {\frac{{{x^2}}}{2}} \right)\,\left( {\frac{{dx}}{{x\sqrt {{x^2} - 1} }}} \right)} \hfill \\ \hfill \\ rewrite \hfill \\ \hfill \\ = \frac{{{x^2}{{\sec }^{ - 1}}x}}{2} - \frac{1}{4}\int_{}^{} {\frac{{2x}}{{\sqrt {{x^2} - 1} }}} \hfill \\ \hfill \\ integrate\,\,{\text{using }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C} \hfill \\ \hfill \\ = \frac{{{x^2}{{\sec }^{ - 1}}x}}{2} - \frac{1}{4}\,\left( {\frac{{\sqrt {{x^2} - 1} }}{{\frac{1}{2}}}} \right) + C \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \frac{{{x^2}{{\sec }^{ - 1}}x}}{2} - \frac{1}{2}\sqrt {{x^2} - 1} + C \hfill \\ \end{gathered}$