$\text{If}$
$\alpha ={\int }_{1/2}^2\frac{{\tan }^{-1}x}{2x^2-3x+2}dx,$
$\text{then the value of }\sqrt{7}\tan \left(\frac{2\alpha \sqrt{7}}{\pi }\right)\text{ is}\text{\_\_\_\_\_\_\_\_}$

$\text{(Here, the inverse trigonometric function}{\tan }^{-1}x\text{assumes values in}(-\frac{\pi }{2},\frac{\pi }{2}).$

The correct answer is 21.

To find the value of the given expression, we start with the definite integral:
$\alpha ={\int }_{1/2}^2\frac{{\tan }^{-1}x}{2x^2-3x+2}dx$

Step 1: Simplify the Integral using Substitution
Let us use the substitution $x=\frac{1}{t}$, which gives $dx=-\frac{1}{t^2}dt$.
We also change the limits of integration:
When $x=\frac{1}{2}$, we have $t=2$.
When $x=2$, we have $t=\frac{1}{2}$.
Substituting these into the integral, we get:
$\alpha ={\int }_2^{1/2}\frac{{\tan }^{-1}(1/t)}{2(1/t^2)-3(1/t)+2}(-\frac{1}{t^2})dt$

Using the negative sign to swap the limits of integration back, we obtain:
$\alpha ={\int }_{1/2}^2\frac{{\tan }^{-1}(1/t)}{t^2\left(\frac{2-3t+2t^2}{t^2}\right)}dt$
Simplifying the denominator:
$\alpha ={\int }_{1/2}^2\frac{{\tan }^{-1}(1/t)}{2t^2-3t+2}dt$

Since the variable of integration is a dummy variable, we can replace $t$ back with $x$:
$\alpha ={\int }_{1/2}^2\frac{{\tan }^{-1}(1/x)}{2x^2-3x+2}dx$

Step 2: Combine the Two Expressions for α
For $x>0$, we have the identity ${\tan }^{-1}(1/x)=\frac{\pi }{2}-{\tan }^{-1}x$. Since the interval of integration $[1/2,2]$ is strictly positive, we substitute this identity:
$\alpha ={\int }_{1/2}^2\frac{\frac{\pi }{2}-{\tan }^{-1}x}{2x^2-3x+2}dx$
Adding this equation to the original definition of $\alpha$:
$2\alpha ={\int }_{1/2}^2\frac{{\tan }^{-1}x+\frac{\pi }{2}-{\tan }^{-1}x}{2x^2-3x+2}dx$
This simplifies to:
$2\alpha =\frac{\pi }{2}{\int }_{1/2}^2\frac{1}{2x^2-3x+2}dx$
Dividing by 2, we get:
$\alpha =\frac{\pi }{4}{\int }_{1/2}^2\frac{1}{2x^2-3x+2}dx$

Step 3: Evaluate the Integral
Let us evaluate $J={\int }_{1/2}^2\frac{1}{2x^2-3x+2}dx$ by completing the square in the denominator:
$2x^2-3x+2=2[x^2-\frac{3}{2}x+1]=2[{(x-\frac{3}{4})}^2+1-\frac{9}{16}]=2[{(x-\frac{3}{4})}^2+\frac{7}{16}]$
Therefore, the integral is:
$J=\frac{1}{2}{\int }_{1/2}^2\frac{1}{{(x-\frac{3}{4})}^2+{\left(\frac{\sqrt{7}}{4}\right)}^2}dx$

Using the standard integral formula $\int \frac{1}{u^2+a^2}du=\frac{1}{a}{\tan }^{-1}\left(\frac{u}{a}\right)$, we have:
$J=\frac{1}{2}[\frac{4}{\sqrt{7}}{\tan }^{-1}\left(\frac{x-3/4}{\sqrt{7}/4}\right){]}_{1/2}^2$
$J=\frac{2}{\sqrt{7}}[{\tan }^{-1}\left(\frac{4x-3/}{\sqrt{7}}\right){]}_{1/2}^2$

Now evaluate this expression at the limits $x=2$ and $x=\frac{1}{2}$:
At $x=2$:
$\frac{4\left(2\right)-3}{\sqrt{7}}=\frac{5}{\sqrt{7}}$
At $x=1/2$:
$\frac{4(1/2)-3}{\sqrt{7}}=-\frac{1}{\sqrt{7}}$

Substituting these values in:
$J=\frac{2}{\sqrt{7}}\left({\tan }^{-1}\right(\frac{5}{\sqrt{7}})-{\tan }^{-1}(-\frac{1}{\sqrt{7}}\left)\right)$
Since ${\tan }^{-1}(-\theta )=-{\tan }^{-1}\theta$:
$J=\frac{2}{\sqrt{7}}\left({\tan }^{-1}\right(\frac{5}{\sqrt{7}})+{\tan }^{-1}(\frac{1}{\sqrt{7}}\left)\right)$

Applying the identity ${\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left(\frac{A+B/}{1-AB}\right)$:
${\tan }^{-1}\left(\frac{5}{\sqrt{7}}\right)+{\tan }^{-1}\left(\frac{1}{\sqrt{7}}\right)={\tan }^{-1}\left(\frac{\frac{5}{\sqrt{7}}+\frac{1}{\sqrt{7}}}{1-\frac{5}{7}}\right)={\tan }^{-1}\left(\frac{6/\sqrt{7}}{2/7}\right)={\tan }^{-1}\left(3\sqrt{7}\right)$

Therefore:
$J=\frac{2}{\sqrt{7}}{\tan }^{-1}\left(3\sqrt{7}\right)$

This gives the value of $\alpha$:
$\alpha =\frac{\pi }{4}J=\frac{\pi }{4}\cdot \frac{2}{\sqrt{7}}{\tan }^{-1}\left(3\sqrt{7}\right)=\frac{\pi }{2\sqrt{7}}{\tan }^{-1}\left(3\sqrt{7}\right)$

Step 4: Compute the Target Expression
We need to find the value of:
$\sqrt{7}\tan \left(\frac{2\alpha \sqrt{7}}{\pi }\right)$
First, substitute the expression for $\alpha$ inside the argument of $\tan$:
$\frac{2\alpha \sqrt{7}}{\pi }=\frac{2\sqrt{7}}{\pi }\left(\frac{\pi }{2\sqrt{7}}{\tan }^{-1}\right(3\sqrt{7}\left)\right)={\tan }^{-1}\left(3\sqrt{7}\right)$

Now, evaluate the final expression:
$\sqrt{7}\tan \left({\tan }^{-1}\right(3\sqrt{7}\left)\right)=\sqrt{7}\cdot 3\sqrt{7}=3\cdot 7=21$