Question 16

Let $I = \alpha = \int_{1/2}^2 \frac{\tan^{-1} x}{2x^2 - 3x + 2} dx$.

Substitute $x = 1/t$, $dx = -1/t^2 dt$:

$I = \int_2^{1/2} \frac{\tan^{-1}(1/t)}{2(1/t)^2 - 3(1/t) + 2} \left(-\frac{1}{t^2}\right) dt = \int_{1/2}^2 \frac{\cot^{-1} t}{2 - 3t + 2t^2} dt$

Adding the two expressions for $I$:

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

$\alpha = \frac{\pi}{8} \int_{1/2}^2 \frac{1}{(x-3/4)^2 + 7/16} dx = \frac{\pi}{8} \left[ \frac{4}{\sqrt{7}} \tan^{-1} \left( \frac{4x-3}{\sqrt{7}} \right) \right]_{1/2}^2$

$\alpha = \frac{\pi}{2\sqrt{7}} [ \tan^{-1}(5/\sqrt{7}) - \tan^{-1}(-1/\sqrt{7}) ] = \frac{\pi}{2\sqrt{7}} \tan^{-1} \left( \frac{6/\sqrt{7}}{1-5/7} \right) = \frac{\pi}{2\sqrt{7}} \tan^{-1}(3\sqrt{7})$

Then $\frac{2\alpha\sqrt{7}}{\pi} = \tan^{-1}(3\sqrt{7})$,

so $\tan \left( \frac{2\alpha \sqrt{7}}{\pi} \right) = 3\sqrt{7}$.

$\sqrt{7} \tan \left( \frac{2\alpha \sqrt{7}}{\pi} \right) = \sqrt{7}(3\sqrt{7}) = 21$.