Answer
$$ - \frac{\pi }{{12}}$$
Work Step by Step
$$\eqalign{
& \int_{ - 2}^{ - 2/\sqrt 3 } {\frac{{dx}}{{x\sqrt {{x^2} - 1} }}} \cr
& {\text{integrate using the formula }}\int {\frac{1}{{\left| u \right|\sqrt {{u^2} - 1} }} = {{\sec }^{ - 1}}u + C,{\text{ }}} \left( {{\text{see page 468}}} \right) \cr
& {\text{let }}a = 1 \cr
& \int_{ - \sqrt 2 }^{ - 2/\sqrt 3 } {\frac{{dx}}{{x\sqrt {{x^2} - 1} }}} = \left. {\left( {{{\sec }^{ - 1}}u} \right)} \right|_{ - \sqrt 2 }^{ - 2/\sqrt 3 } \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = {\sec ^{ - 1}}\left( { - \frac{2}{{\sqrt 2 }}} \right) - {\sec ^{ - 1}}\left( { - \sqrt 2 } \right) \cr
& or \cr
& = co{s^{ - 1}}\left( { - \frac{{\sqrt 3 }}{2}} \right) - {\cos ^{ - 1}}\left( { - \frac{1}{{\sqrt 2 }}} \right) \cr
& {\text{simplify}} \cr
& = \frac{{5\pi }}{6} - \frac{{3\pi }}{4} \cr
& = \frac{{10\pi - 9\pi }}{{12}} \cr
& = - \frac{\pi }{{12}} \cr} $$