Answer
$$\frac{{5\pi - 3\sqrt 3 }}{9}$$
Work Step by Step
$$\eqalign{
& \int_{2/\sqrt 3 }^2 {t{{\sec }^{ - 1}}t} dt \cr
& {\text{Using integration by parts method }} \cr
& \,\,\,\,\,{\text{Let }}u = {\sec ^{ - 1}}t,\,\,\,\,du = \frac{1}{{t\sqrt {{t^2} - 1} }}dt,\,\,\,\,\,dv = tdt,\,\,\,\,\,v = \frac{1}{2}{t^2} \cr
& \cr
& {\text{Integration by parts then gives}} \cr
& \int {t{{\sec }^{ - 1}}t} dt = \frac{1}{2}{t^2}{\sec ^{ - 1}}t - \int {\left( {\frac{1}{2}{t^2}} \right)} \left( {\frac{1}{{t\sqrt {{t^2} - 1} }}dt} \right) \cr
& \int {t{{\sec }^{ - 1}}t} dt = \frac{1}{2}{t^2}{\sec ^{ - 1}}t - \frac{1}{4}\int {\frac{{2t}}{{\sqrt {{t^2} - 1} }}dt} \cr
& = \frac{1}{2}{t^2}{\sec ^{ - 1}}t - \frac{1}{4}\int {{{\left( {{t^2} - 1} \right)}^{ - 1/2}}\left( {2t} \right)dt} \cr
& = \frac{1}{2}{t^2}{\sec ^{ - 1}}t - \frac{1}{4}\int {{{\left( {{t^2} - 1} \right)}^{ - 1/2}}\left( {2t} \right)dt} \cr
& = \frac{1}{2}{t^2}{\sec ^{ - 1}}t - \frac{1}{4}\left( {\frac{{{{\left( {{t^2} - 1} \right)}^{1/2}}}}{{1/2}}} \right) + C \cr
& = \frac{1}{2}{t^2}{\sec ^{ - 1}}t - \frac{1}{2}\sqrt {{t^2} - 1} + C \cr
& {\text{Then}}{\text{,}} \cr
& \int_{2/\sqrt 3 }^2 {t{{\sec }^{ - 1}}t} dt = \left( {\frac{1}{2}{t^2}{{\sec }^{ - 1}}t - \frac{1}{2}\sqrt {{t^2} - 1} } \right)_{2/\sqrt 3 }^2 \cr
& = \left( {\frac{1}{2}{{\left( 2 \right)}^2}{{\sec }^{ - 1}}\left( 2 \right) - \frac{1}{2}\sqrt {{2^2} - 1} } \right) - \left( {\frac{1}{2}{{\left( {\frac{2}{{\sqrt 3 }}} \right)}^2}{{\sec }^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right) - \frac{1}{2}\sqrt {{{\left( {\frac{2}{{\sqrt 3 }}} \right)}^2} - 1} } \right) \cr
& = \left( {\frac{2}{3}\pi - \frac{{\sqrt 3 }}{2}} \right) - \left( {\frac{1}{9}\pi - \frac{{\sqrt 3 }}{6}} \right) \cr
& = \frac{2}{3}\pi - \frac{{\sqrt 3 }}{2} - \frac{1}{9}\pi + \frac{{\sqrt 3 }}{6} \cr
& = \frac{5}{9}\pi - \frac{{\sqrt 3 }}{3} \cr
& = \frac{{5\pi - 3\sqrt 3 }}{9} \cr} $$
