If $\alpha = \int_{\frac{1}{2}}^{2} \frac{\tan^{-1} x}{2x^{2} - 3x + 2}\,dx$, 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 $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$).

If α = ∫_(1/2)² (tan⁻¹ x)/(2x² - 3x + 2) dx, then the value of √7tan(2α√7/π) is ________. (Here, the inverse trigonometric function tan⁻¹ x assumes values in (-π/2, π/2)). $\alpha = \int_{1/2}^{2} \frac{\tan^{-1} x}{2x^2 - 3x + 2}\,dx$ $\tan^{-1} x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ $\text{Find: } \sqrt{7}\tan\!\left(\frac{2\alpha\sqrt{7}}{\pi}\right)$ $x = \frac{1}{t} \Rightarrow 2x^2-3x+2 = \frac{2t^2-3t+2}{t^2}$ $I = \int_{1/2}^{2} \frac{\tan^{-1}\frac{1}{x}}{2x^2-3x+2}\,dx$ $2I = \int_{1/2}^{2} \frac{\tan^{-1} x + \tan^{-1}\frac{1}{x}}{2x^2-3x+2}\,dx$ $\tan^{-1} x + \tan^{-1}\frac{1}{x} = \frac{\pi}{2} \quad (x > 0)$ $2I = \frac{\pi}{2}\int_{1/2}^{2} \frac{dx}{2x^2-3x+2}$ $I = \frac{\pi}{4}\int_{1/2}^{2} \frac{dx}{2x^2-3x+2}$ $2x^2-3x+2 = 2\left[\left(x-\frac{3}{4}\right)^2 + \frac{7}{16}\right]$ $\int\frac{dx}{2x^2-3x+2} = \frac{2}{\sqrt{7}}\left[\tan^{-1}\frac{4x-3}{\sqrt{7}}\right]_{1/2}^{2}$ $= \frac{2}{\sqrt{7}}\left[\tan^{-1}\frac{5}{\sqrt{7}} + \tan^{-1}\frac{1}{\sqrt{7}}\right]$ $\tan^{-1}\frac{5}{\sqrt{7}} + \tan^{-1}\frac{1}{\sqrt{7}} = \tan^{-1}(3\sqrt{7})$ $I = \alpha = \frac{\pi}{4}\cdot\frac{2}{\sqrt{7}}\tan^{-1}(3\sqrt{7}) = \frac{\pi}{2\sqrt{7}}\tan^{-1}(3\sqrt{7})$ $\frac{2\alpha\sqrt{7}}{\pi} = \tan^{-1}(3\sqrt{7})$ $\sqrt{7}\tan\!\left(\frac{2\alpha\sqrt{7}}{\pi}\right) = \sqrt{7}\cdot 3\sqrt{7} = 21$