Value of Definite Integral Involving Arctan and Quadratic Denominator

To find the value of $\sqrt{7} \tan\left( \frac{2\alpha \sqrt{7}}{\pi} \right)$, we begin by simplifying the given definite integral:

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

Using the substitution $x = \frac{1}{t}$, we have $dx = -\frac{1}{t^2} \, dt$.

The limits of integration change as follows:

• When $x = \frac{1}{2}$, $t = 2$.
• When $x = 2$, $t = \frac{1}{2}$.

Substituting these into equation (1): $\alpha = \int_{2}^{1/2} \frac{\tan^{-1}\left(\frac{1}{t}\right)}{2\left(\frac{1}{t}\right)^2 - 3\left(\frac{1}{t}\right) + 2} \left( -\frac{1}{t^2} \right) dt$

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

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

Replacing the dummy variable $t$ back with $x$: $\alpha = \int_{1/2}^{2} \frac{\tan^{-1}\left(\frac{1}{x}\right)}{2x^2 - 3x + 2} \, dx \quad \text{--- (2)}$

Adding equation (1) and equation (2): $2\alpha = \int_{1/2}^{2} \frac{\tan^{-1} x + \tan^{-1}\left(\frac{1}{x}\right)}{2x^2 - 3x + 2} \, dx$

Since $x > 0$ for $x \in \left[\frac{1}{2}, 2\right]$, we use the identity $\tan^{-1} x + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2}$: $2\alpha = \int_{1/2}^{2} \frac{\frac{\pi}{2}}{2x^2 - 3x + 2} \, dx$

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

Now, we evaluate the remaining integral $I$: $I = \int_{1/2}^{2} \frac{dx}{2x^2 - 3x + 2}$

Completing the square in the denominator: $2x^2 - 3x + 2 = 2\left(x^2 - \frac{3}{2}x + 1\right) = 2\left[\left(x - \frac{3}{4}\right)^2 + 1 - \frac{9}{16}\right] = 2\left[\left(x - \frac{3}{4}\right)^2 + \frac{7}{16}\right]$

Thus, $I = \frac{1}{2} \int_{1/2}^{2} \frac{dx}{\left(x - \frac{3}{4}\right)^2 + \left(\frac{\sqrt{7}}{4}\right)^2}$

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

$I = \frac{2}{\sqrt{7}} \left[ \tan^{-1}\left(\frac{4x - 3}{\sqrt{7}}\right) \right]_{1/2}^2$

Evaluating at the limits: $I = \frac{2}{\sqrt{7}} \left[ \tan^{-1}\left(\frac{5}{\sqrt{7}}\right) - \tan^{-1}\left(\frac{-1}{\sqrt{7}}\right) \right]$

$I = \frac{2}{\sqrt{7}} \left[ \tan^{-1}\left(\frac{5}{\sqrt{7}}\right) + \tan^{-1}\left(\frac{1}{\sqrt{7}}\right) \right]$

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

Hence, $I = \frac{2}{\sqrt{7}} \tan^{-1}(3\sqrt{7})$

Substitute $I$ back into the expression for $\alpha$: $\alpha = \frac{\pi}{4} \cdot \frac{2}{\sqrt{7}} \tan^{-1}(3\sqrt{7}) = \frac{\pi}{2\sqrt{7}} \tan^{-1}(3\sqrt{7})$

Rearranging terms: $\frac{2\alpha \sqrt{7}}{\pi} = \tan^{-1}(3\sqrt{7})$

Taking the tangent on both sides: $\tan\left( \frac{2\alpha \sqrt{7}}{\pi} \right) = 3\sqrt{7}$

Multiplying both sides by $\sqrt{7}$: $\sqrt{7} \tan\left( \frac{2\alpha \sqrt{7}}{\pi} \right) = \sqrt{7} \times 3\sqrt{7} = 21$