Question:

Let $x = 1/t \implies dx = -1/t^2 dt$. Change the limits from $[1/2, 2]$ to $[2, 1/2]$: $$\alpha = \int_{2}^{1/2} \dfrac{\tan^{-1}(1/t)}{2/t^2 - 3/t + 2} \left(-\dfrac{1}{t^2}\right) dt = \int_{1/2}^2 \dfrac{\cot^{-1} t}{2t^2 - 3t + 2} dt$$ Using dummy variable $x$: $$\alpha = \int_{1/2}^2 \dfrac{\cot^{-1} x}{2x^2 - 3x + 2} dx$$ Add the two forms of $\alpha$: $$2\alpha = \int_{1/2}^2 \dfrac{\tan^{-1} x + \cot^{-1} x}{2x^2 - 3x + 2} dx$$ Since $\tan^{-1} x + \cot^{-1} x = \pi/2$ for all $x > 0$: $$2\alpha = \dfrac{\pi}{2} \int_{1/2}^2 \dfrac{1}{2x^2 - 3x + 2} dx \implies \alpha = \dfrac{\pi}{4} \int_{1/2}^2 \dfrac{1}{2x^2 - 3x + 2} dx$$ Complete the square of the denominator: $$2x^2 - 3x + 2 = 2\left[\left(x - \dfrac{3}{4}\right)^2 + \\dfrac{7}{16}\right]$$ Integrate: $$\int_{1/2}^2 \dfrac{1}{2x^2 - 3x + 2} dx = \dfrac{1}{2} \cdot \dfrac{4}{\sqrt{7}} \left[ \tan^{-1}\left(\dfrac{4x - 3}{\sqrt{7}}\right) \right]_{1/2}^2$$ $$= \dfrac{2}{\sqrt{7}} \left[ \tan^{-1}\left(\dfrac{5}{\sqrt{7}}\right) - \tan^{-1}\left(-\dfrac{1}{\sqrt{7}}\right) \right]$$ Use $\tan^{-1} A + \tan^{-1} B = \tan^{-1}\left(\dfrac{A+B}{1-AB}\right)$: $$\tan^{-1}\left(\dfrac{5}{\sqrt{7}}\right) + \tan^{-1}\left(\dfrac{1}{\sqrt{7}}\right) = \tan^{-1}\left(\dfrac{6/\sqrt{7}}{2/7}\right) = \tan^{-1}(3\sqrt{7})$$ So the integral evaluates to: $$\alpha = \dfrac{\pi}{4} \cdot \dfrac{2}{\sqrt{7}} \tan^{-1}(3\sqrt{7}) = \dfrac{\pi}{2\sqrt{7}} \tan^{-1}(3\sqrt{7})$$ Rearranging gives: $$\dfrac{2\alpha\sqrt{7}}{\pi} = \tan^{-1}(3\sqrt{7}) \implies \tan\left(\dfrac{2\alpha\sqrt{7}}{\pi}\right) = 3\sqrt{7}$$ Thus: $\sqrt{7} \tan\left(\dfrac{2\alpha\sqrt{7}}{\pi}\right) = \sqrt{7}(3\sqrt{7}) = 21$.