Step-by-step Solution:

1. Analyze the Integral and Apply Substitution

The given integral is: $\alpha=\int\limits_{\frac{1}{2}}^2 \frac{\tan ^{-1} x}{2 x^2-3 x+2} d x \quad \dots (1)$ The limits of integration are $1/2$ and $2$, which are reciprocals of each other. This suggests using the substitution $x = 1/t$. If $x = 1/t$, then $dx = -1/t^2 dt$. The limits of integration transform as follows:

• When $x = 1/2$, $t = 2$.
• When $x = 2$, $t = 1/2$.

Substituting into the integral: $\alpha = \int_2^{1/2} \frac{\tan^{-1}(1/t)}{2(1/t)^2 - 3(1/t) + 2} \left(-\frac{1}{t^2}\right) dt$ Simplify the denominator: $2(1/t)^2 - 3(1/t) + 2 = \frac{2 - 3t + 2t^2}{t^2}$ Now, substitute this back into the integral expression: $\alpha = \int_2^{1/2} \frac{\tan^{-1}(1/t)}{\frac{2t^2 - 3t + 2}{t^2}} \left(-\frac{1}{t^2}\right) dt = \int_2^{1/2} \frac{-\tan^{-1}(1/t)}{2t^2 - 3t + 2} dt$ Using the property $\int_a^b f(x) dx = -\int_b^a f(x) dx$, we can flip the limits: $\alpha = \int_{1/2}^2 \frac{\tan^{-1}(1/t)}{2t^2 - 3t + 2} dt$ For $t > 0$, we have the identity $\tan^{-1}(1/t) = \frac{\pi}{2} - \tan^{-1}(t)$. Since the integration is over $[1/2, 2]$, $t$ is positive. $\alpha = \int_{1/2}^2 \frac{\frac{\pi}{2} - \tan^{-1}(t)}{2t^2 - 3t + 2} dt$ Since the variable of integration is a dummy variable, we can replace $t$ with $x$: $\alpha = \int_{1/2}^2 \frac{\frac{\pi}{2} - \tan^{-1}(x)}{2x^2 - 3x + 2} dx \quad \dots (2)$

2. Combine the Expressions for $\alpha$

Add equation (1) and equation (2): $\alpha + \alpha = \int_{1/2}^2 \frac{\tan^{-1} x}{2x^2 - 3x + 2} dx + \int_{1/2}^2 \frac{\frac{\pi}{2} - \tan^{-1}(x)}{2x^2 - 3x + 2} dx$ $2\alpha = \int_{1/2}^2 \frac{\tan^{-1} x + \frac{\pi}{2} - \tan^{-1}(x)}{2x^2 - 3x + 2} dx$ $2\alpha = \int_{1/2}^2 \frac{\frac{\pi}{2}}{2x^2 - 3x + 2} dx = \frac{\pi}{2} \int_{1/2}^2 \frac{1}{2x^2 - 3x + 2} dx$

3. Evaluate the Simplified Integral

Let's evaluate the integral $I = \int_{1/2}^2 \frac{1}{2x^2 - 3x + 2} dx$. We complete the square for the denominator: $2x^2 - 3x + 2 = 2\left(x^2 - \frac{3}{2}x + 1\right) = 2\left(\left(x - \frac{3}{4}\right)^2 - \frac{9}{16} + 1\right) = 2\left(\left(x - \frac{3}{4}\right)^2 + \frac{7}{16}\right)$ The integral becomes: $I = \int_{1/2}^2 \frac{1}{2\left(\left(x - \frac{3}{4}\right)^2 + \left(\frac{\sqrt{7}}{4}\right)^2\right)} dx = \frac{1}{2} \int_{1/2}^2 \frac{1}{\left(x - \frac{3}{4}\right)^2 + \left(\frac{\sqrt{7}}{4}\right)^2} dx$ This is a standard integral of the form $\int \frac{1}{u^2+a^2} du = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right)$. $I = \frac{1}{2} \left[ \frac{1}{\frac{\sqrt{7}}{4}} \tan^{-1}\left(\frac{x - \frac{3}{4}}{\frac{\sqrt{7}}{4}}\right) \right]_{1/2}^2 = \frac{1}{2} \cdot \frac{4}{\sqrt{7}} \left[ \tan^{-1}\left(\frac{4x - 3}{\sqrt{7}}\right) \right]_{1/2}^2$ $I = \frac{2}{\sqrt{7}} \left[ \tan^{-1}\left(\frac{4(2) - 3}{\sqrt{7}}\right) - \tan^{-1}\left(\frac{4(1/2) - 3}{\sqrt{7}}\right) \right]$ $I = \frac{2}{\sqrt{7}} \left[ \tan^{-1}\left(\frac{5}{\sqrt{7}}\right) - \tan^{-1}\left(\frac{-1}{\sqrt{7}}\right) \right] = \frac{2}{\sqrt{7}} \left[ \tan^{-1}\left(\frac{5}{\sqrt{7}}\right) + \tan^{-1}\left(\frac{1}{\sqrt{7}}\right) \right]$ Using the identity $\tan^{-1} A + \tan^{-1} B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$ (since $AB = \frac{5}{7} < 1$): $I = \frac{2}{\sqrt{7}} \tan^{-1}\left(\frac{\frac{5}{\sqrt{7}} + \frac{1}{\sqrt{7}}}{1 - \frac{5}{\sqrt{7}} \cdot \frac{1}{\sqrt{7}}}\right) = \frac{2}{\sqrt{7}} \tan^{-1}\left(\frac{\frac{6}{\sqrt{7}}}{1 - \frac{5}{7}}\right) = \frac{2}{\sqrt{7}} \tan^{-1}\left(\frac{\frac{6}{\sqrt{7}}}{\frac{2}{7}}\right)$ $I = \frac{2}{\sqrt{7}} \tan^{-1}\left(\frac{6}{\sqrt{7}} \cdot \frac{7}{2}\right) = \frac{2}{\sqrt{7}} \tan^{-1}(3\sqrt{7})$

4. Find the Value of $\alpha$

Substitute the value of $I$ back into the expression for $2\alpha$: $2\alpha = \frac{\pi}{2} I = \frac{\pi}{2} \cdot \frac{2}{\sqrt{7}} \tan^{-1}(3\sqrt{7}) = \frac{\pi}{\sqrt{7}} \tan^{-1}(3\sqrt{7})$

5. Calculate the Final Expression

We need to find the value of $\sqrt{7} \tan \left(\frac{2 \alpha \sqrt{7}}{\pi}\right)$. From the expression for $2\alpha$, we can find the argument of the tangent function: $\frac{2 \alpha \sqrt{7}}{\pi} = \frac{\sqrt{7}}{\pi} \left( \frac{\pi}{\sqrt{7}} \tan^{-1}(3\sqrt{7}) \right) = \tan^{-1}(3\sqrt{7})$ Now, substitute this into the final expression: $\sqrt{7} \tan \left(\frac{2 \alpha \sqrt{7}}{\pi}\right) = \sqrt{7} \tan\left(\tan^{-1}(3\sqrt{7})\right)$ Using the property $\tan(\tan^{-1}(y)) = y$: $= \sqrt{7} \cdot (3\sqrt{7}) = 3 \cdot (\sqrt{7})^2 = 3 \cdot 7 = 21$

The value of the expression is 21.