Answer
\[\frac{2}{3}\pi - \frac{1}{2}\sqrt 3 \]
Work Step by Step
\[\begin{gathered}
\int_1^2 {x{{\sec }^{ - 1}}x} dx \hfill \\
{\text{Let }}u = {\sec ^{ - 1}}x{\text{ and }}dv = xdx \hfill \\
du = \frac{1}{{x\sqrt {{x^2} - 1} }}dx{\text{ }}v = \frac{{{x^2}}}{2} \hfill \\
{\text{Use integration by parts formula}} \hfill \\
\int {udv} = uv - \int {vdu} \hfill \\
\int {x{{\sec }^{ - 1}}xdx} = \frac{{{x^2}}}{2}{\sec ^{ - 1}}x - \int {\frac{1}{{x\sqrt {{x^2} - 1} }}\left( {\frac{{{x^2}}}{2}} \right)} dx \hfill \\
\int {x{{\sec }^{ - 1}}xdx} = \frac{{{x^2}}}{2}{\sec ^{ - 1}}x - \frac{1}{2}\int {\frac{x}{{\sqrt {{x^2} - 1} }}} dx \hfill \\
{\text{ }} = \frac{{{x^2}}}{2}{\sec ^{ - 1}}x - \frac{1}{4}\int {\frac{{2x}}{{\sqrt {{x^2} - 1} }}} dx \hfill \\
{\text{ }} = \frac{{{x^2}}}{2}{\sec ^{ - 1}}x - \frac{1}{4}\left( {2\sqrt {{x^2} - 1} } \right) + C \hfill \\
{\text{ }} = \frac{{{x^2}}}{2}{\sec ^{ - 1}}x - \frac{1}{2}\sqrt {{x^2} - 1} + C \hfill \\
{\text{Therefore,}} \hfill \\
\int_1^2 {x{{\sec }^{ - 1}}x} dx = \left[ {\frac{{{x^2}}}{2}{{\sec }^{ - 1}}x - \frac{1}{2}\sqrt {{x^2} - 1} } \right]_1^2 \hfill \\
= \left[ {\frac{{{2^2}}}{2}{{\sec }^{ - 1}}2 - \frac{1}{2}\sqrt {{2^2} - 1} } \right] - \left[ {\frac{{{1^2}}}{2}{{\sec }^{ - 1}}1 - \frac{1}{2}\sqrt {{1^2} - 1} } \right] \hfill \\
{\text{Simplify}} \hfill \\
= \left[ {2\left( {\frac{\pi }{3}} \right) - \frac{1}{2}\sqrt 3 } \right] - \left[ {\frac{{{1^2}}}{2}\left( 0 \right) - \frac{1}{2}\sqrt 0 } \right] \hfill \\
= \frac{2}{3}\pi - \frac{1}{2}\sqrt 3 \hfill \\
\end{gathered} \]
