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

The given integral is$$\alpha=\int_{1/2}^{2}\frac{\tan^{-1}x}{2x^{2}-3x+2}\,dx.$$

Step 1 : Use the reciprocal-substitution symmetry.
Put $$x=\frac{1}{t}$$ so that $$dx=-\frac{dt}{t^{2}}.$$ The limits change as $$x : 1/2 \to 2$$ implies $$t : 2 \to 1/2$$.

Then

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

Step 2 : Add the two forms of the integral.

Adding the original and the transformed integrals,

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

For every positive $$x\neq1$$, $$\tan^{-1}x+\tan^{-1}\!\left(\frac1x\right)=\frac{\pi}{2}.$$ (The value at $$x=1$$ does not affect the integral.)

Hence

$$2\alpha=\frac{\pi}{2}\int_{1/2}^{2}\frac{dx}{2x^{2}-3x+2},\qquad \alpha=\frac{\pi}{4}\int_{1/2}^{2}\frac{dx}{2x^{2}-3x+2}.$$

Step 3 : Evaluate the remaining integral.

Rewrite the quadratic:

$$2x^{2}-3x+2=2\!\left[x^{2}-\frac{3}{2}x+1\right] =2\!\Bigl[(x-\tfrac34)^{2}+\tfrac7{16}\Bigr] =2\!\Bigl[(x-\tfrac34)^{2}+(\tfrac{\sqrt7}{4})^{2}\Bigr].$$

Therefore

$$\int_{1/2}^{2}\frac{dx}{2x^{2}-3x+2} =\frac12\int_{1/2}^{2}\frac{dx}{(x-\tfrac34)^{2}+(\tfrac{\sqrt7}{4})^{2}} =\frac12\cdot\frac4{\sqrt7}\Bigl[\tan^{-1}\!\bigl(\tfrac{4x-3}{\sqrt7}\bigr)\Bigr]_{1/2}^{2}.$$

Simplifying the constant factor gives $$\dfrac{2}{\sqrt7}$$. Now compute the bracketed values:

At $$x=2$$: $$\dfrac{4x-3}{\sqrt7}=\dfrac{5}{\sqrt7}$$.
At $$x=\tfrac12$$: $$\dfrac{4x-3}{\sqrt7}=-\dfrac1{\sqrt7}$$.

Hence

$$\int_{1/2}^{2}\frac{dx}{2x^{2}-3x+2} =\frac{2}{\sqrt7}\Bigl[\tan^{-1}\!\bigl(\tfrac{5}{\sqrt7}\bigr)+\tan^{-1}\!\bigl(\tfrac{1}{\sqrt7}\bigr)\Bigr].$$

Using the addition formula $$\tan^{-1}a+\tan^{-1}b=\tan^{-1}\!\bigl(\dfrac{a+b}{1-ab}\bigr)$$ (since $$ab\lt 1$$),

$$\tan^{-1}\!\bigl(\tfrac{5}{\sqrt7}\bigr)+\tan^{-1}\!\bigl(\tfrac{1}{\sqrt7}\bigr) =\tan^{-1}\!\Bigl(\frac{\,\tfrac{5}{\sqrt7}+\tfrac{1}{\sqrt7}}{1-\tfrac{5}{7}}\Bigr) =\tan^{-1}\!\Bigl(\frac{\,\tfrac{6}{\sqrt7}}{\tfrac{2}{7}}\Bigr) =\tan^{-1}\!\Bigl(\frac{21}{\sqrt7}\Bigr).$$

Thus

$$\int_{1/2}^{2}\frac{dx}{2x^{2}-3x+2} =\frac{2}{\sqrt7}\,\tan^{-1}\!\Bigl(\frac{21}{\sqrt7}\Bigr).$$

Step 4 : Obtain $$\alpha$$.

$$\alpha=\frac{\pi}{4}\times\frac{2}{\sqrt7}\, \tan^{-1}\!\Bigl(\frac{21}{\sqrt7}\Bigr) =\frac{\pi}{2\sqrt7}\, \tan^{-1}\!\Bigl(\frac{21}{\sqrt7}\Bigr).$$

Step 5 : Evaluate the required expression.

Compute the angle inside the tangent:

$$\frac{2\alpha\sqrt7}{\pi} =\frac{2}{\pi}\cdot\frac{\pi}{2\sqrt7}\,\tan^{-1}\!\Bigl(\frac{21}{\sqrt7}\Bigr)\!\cdot\sqrt7 =\tan^{-1}\!\Bigl(\frac{21}{\sqrt7}\Bigr).$$

Therefore

$$\sqrt7\,\tan\!\Bigl(\frac{2\alpha\sqrt7}{\pi}\Bigr) =\sqrt7\,\tan\!\left[\tan^{-1}\!\Bigl(\frac{21}{\sqrt7}\Bigr)\right] =\sqrt7\cdot\frac{21}{\sqrt7}=21.$$

Hence, the required value is 21.