Mathematics - Definite Integration Question with Solution | TestHub

$\alpha ={\int }_{\frac{1}{2}}^2\frac{{\tan }^{-1}x}{2x^2-3x+2}dx$ ....(i) $\begin{array}{rl}& \text{ Let }x=\frac{1}{t}\\ & dx=-\frac{1}{t^2}dt\\ & \alpha ={\int }_2^{\frac{1}{2}}\frac{{\tan }^{-1}\left(\frac{1}{t}\right)}{\frac{2}{t^2}-\frac{3}{t}+2}\left(\frac{-1}{t^2}\right)dt\end{array}$ $\alpha ={\int }_{\frac{1}{2}}^2\frac{{\cot }^{-1}\mathrm{t}}{2{\mathrm{t}}^2-3\mathrm{t}+2}\mathrm{dt}$ ....(ii) Now by (i) + (ii) $\begin{array}{rl}& 2\alpha ={\int }_{\frac{1}{2}}^2\frac{\frac{\pi }{2}}{2x^2-3x+2}dx\\ & \alpha =\frac{\pi }{8}{\int }_{\frac{1}{2}}^2\frac{dx}{x^2-\frac{3x}{2}+1}\\ & \alpha =\frac{\pi }{8}{\int }_{\frac{1}{2}}^2{\left(x-\frac{3}{4}\right)}^2+\frac{7}{16}\\ & \alpha =\frac{\pi }{8\times \frac{\sqrt{7}}{4}}{\left[{\tan }^{-1}\left(\frac{x-\frac{3}{4}}{\frac{\sqrt{7}}{4}}\right)\right]}_{\frac{1}{2}}^2\end{array}$ $\begin{array}{rl}& \alpha =\frac{\pi }{2\sqrt{7}}{\left[{\tan }^{-1}\frac{4x-3}{\sqrt{7}}\right]}_{\frac{1}{2}}^2\\ & \alpha =\frac{\pi }{2\sqrt{7}}\left[{\tan }^{-1}\frac{5}{\sqrt{7}}-{\tan }^{-1}\left(-\frac{1}{\sqrt{7}}\right)\right]\\ & \alpha =\frac{\pi }{2\sqrt{7}}{\tan }^{-1}\frac{\left(\frac{5}{\sqrt{7}}+\frac{1}{\sqrt{7}}\right)}{1-\frac{5}{7}}\\ & \alpha =\frac{\pi }{2\sqrt{7}}{\tan }^{-1}(3\sqrt{7})\end{array}$ Now $\sqrt{7}\tan \left(\frac{2\sqrt{7}\alpha }{\pi }\right)$ $\begin{array}{rl}& \sqrt{7}\times \tan \left({\tan }^{-1}(3\sqrt{7})\right)\\ & \sqrt{7}\times 3\sqrt{7}\\ & =21\end{array}$