If
$$\alpha = \int _ { \frac { 1 } { 2 } } ^ { 2 } \frac { \tan ^ { - 1 } x } { 2 x ^ { 2 } - 3 x + 2 } d x$$
then the value of $\sqrt { 7 } \tan \left( \frac { 2 \alpha \sqrt { 7 } } { \pi } \right)$ is $\_\_\_\_$.\\
(Here, the inverse trigonometric function $\tan ^ { - 1 } x$ assumes values in $\left( - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right)$.)