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 \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).)

\(\alpha=\int_{\frac{1}{2}}^2 \frac{\tan ^{-1} x}{2 x^2-3 x+2} d x\) ....(i)
\(\begin{aligned} & \text { Let } x=\frac{1}{t} \\ & \qquad d x=-\frac{1}{t^2} d t \\ & \alpha=\int_2^{\frac{1}{2}} \frac{\tan ^{-1}\left(\frac{1}{t}\right)}{\frac{2}{t^2}-\frac{3}{t}+2}\left(\frac{-1}{t^2}\right) d t\end{aligned}\)
\(\alpha=\int_{\frac{1}{2}}^2 \frac{\cot ^{-1} \mathrm{t}}{2 \mathrm{t}^2-3 \mathrm{t}+2} \mathrm{dt}\) ....(ii)
Now by (i) + (ii)
\(\begin{aligned}
& 2 \alpha=\int_{\frac{1}{2}}^2 \frac{\frac{\pi}{2}}{2 x^2-3 x+2} d x \\
& \alpha=\frac{\pi}{8} \int_{\frac{1}{2}}^2 \frac{d x}{x^2-\frac{3 x}{2}+1} \\
& \alpha=\frac{\pi}{8} \int_{\frac{1}{2}}^2\left(x-\frac{3}{4}\right)^2+\frac{7}{16} \\
& \alpha=\frac{\pi}{8 \times \frac{\sqrt{7}}{4}}\left[\tan ^{-1}\left(\frac{x-\frac{3}{4}}{\frac{\sqrt{7}}{4}}\right)\right]_{\frac{1}{2}}^2
\end{aligned}\)
\(\begin{aligned}
& \alpha=\frac{\pi}{2 \sqrt{7}}\left[\tan ^{-1} \frac{4 x-3}{\sqrt{7}}\right]_{\frac{1}{2}}^2 \\
& \alpha=\frac{\pi}{2 \sqrt{7}}\left[\tan ^{-1} \frac{5}{\sqrt{7}}-\tan ^{-1}\left(-\frac{1}{\sqrt{7}}\right)\right] \\
& \alpha=\frac{\pi}{2 \sqrt{7}} \tan ^{-1} \frac{\left(\frac{5}{\sqrt{7}}+\frac{1}{\sqrt{7}}\right)}{1-\frac{5}{7}} \\
& \alpha=\frac{\pi}{2 \sqrt{7}} \tan ^{-1}(3 \sqrt{7})
\end{aligned}\)
Now \(\sqrt{7} \tan \left(\frac{2 \sqrt{7} \alpha}{\pi}\right)\)
\(\begin{aligned}
& \sqrt{7} \times \tan \left(\tan ^{-1}(3 \sqrt{7})\right) \\
& \sqrt{7} \times 3 \sqrt{7} \\
& =21
\end{aligned}\)