JEE Advanced 2025 — Paper 2 — Question 32

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

Let $x=\frac{1}{t}$

$\mathrm{dx}=-\frac{1}{\mathrm{t}^{2}} \mathrm{dt}$

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

$\alpha=\int_{\frac{1}{2}}^{2} \frac{\cot ^{-1} \mathrm{t}}{2 \mathrm{t}^{2}-3 \mathrm{t}+2} \mathrm{dt}$

Now by (i) + (ii)

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

$\alpha=\frac{\pi}{8} \int_{\frac{1}{2}}^{2} \frac{\mathrm{dx}}{\mathrm{x}^{2}-\frac{3 \mathrm{x}}{2}+1}$

$\alpha=\frac{\pi}{8} \int_{\frac{1}{2}}^{2} \frac{\mathrm{dx}}{\left(\mathrm{x}-\frac{3}{4}\right)^{2}+\frac{7}{16}}$

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

$\alpha=\frac{\pi}{2 \sqrt{7}}\left[\tan ^{-1} \frac{4 x-3}{\sqrt{7}}\right]_{\frac{1}{2}}^{2}$

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

$\alpha=\frac{\pi}{2 \sqrt{7}} \tan ^{-1} \frac{\left(\frac{5}{\sqrt{7}}+\frac{1}{\sqrt{7}}\right)}{1-\frac{5}{7}}$

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

Now $\sqrt{7} \tan \left(\frac{2 \sqrt{7} \alpha}{\pi}\right)$

$\sqrt{7} \times \tan \left(\tan ^{-1}(3 \sqrt{7})\right)$

$\sqrt{7} \times 3 \sqrt{7}=21$