Question 16

**Setting up** $$\alpha=\int_{1/2}^{2}\frac{\tan^{-1}x}{2x^2-3x+2}\,dx$$ Substitute $x\to 1/x$. The denominator becomes $\dfrac{2-3x+2x^2}{x^2}$ and $dx\to -\dfrac{dx}{x^2}$, so $$\alpha=\int_{1/2}^{2}\frac{\tan^{-1}(1/x)}{2x^2-3x+2}\,dx .$$ Adding the two forms (since $\tan^{-1}x+\tan^{-1}(1/x)=\frac{\pi}{2}$ for $x>0$): $$2\alpha=\frac{\pi}{2}\int_{1/2}^{2}\frac{dx}{2x^2-3x+2}.$$ **Evaluating the integral** $$2x^2-3x+2 = 2\left[\left(x-\tfrac34\right)^2+\tfrac{7}{16}\right]$$ $$\int\frac{dx}{2x^2-3x+2}=\frac{2}{\sqrt7}\tan^{-1}\!\left(\frac{4x-3}{\sqrt7}\right)$$ From $x=\tfrac12$ to $x=2$: $$\frac{2}{\sqrt7}\left[\tan^{-1}\frac{5}{\sqrt7}+\tan^{-1}\frac{1}{\sqrt7}\right] =\frac{2}{\sqrt7}\tan^{-1}\!\left(\frac{6/\sqrt7}{1-5/7}\right)=\frac{2}{\sqrt7}\tan^{-1}(3\sqrt7).$$ Thus $$2\alpha=\frac{\pi}{\sqrt7}\tan^{-1}(3\sqrt7)\quad\Longrightarrow\quad \frac{2\alpha\sqrt7}{\pi}=\tan^{-1}(3\sqrt7).$$ **Final value** $$\sqrt7\tan\!\left(\frac{2\alpha\sqrt7}{\pi}\right)=\sqrt7\cdot 3\sqrt7=21.$$

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