JEE Main 2026 — 4 April, Morning Shift — Question 42

$\frac{d y}{y^{2}-y+1}=\left(x^{2}+x+1\right) d x$ $\frac{d y}{\left(y-\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}=\left(x^{2}+x+1\right) d x$ $\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{y-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)=\frac{x^{3}}{3}+\frac{x^{3}}{2}+x+C$ $\frac{2}{\sqrt{3 \sqrt{3}}}=\tan ^{-1}\left(\frac{2 y-1}{\sqrt{3}}\right)=\frac{x^{3}}{3}+\frac{x^{3}}{2}+x+C$ $y(0)=\frac{1}{2} \Rightarrow 0=0+0+0+C \Rightarrow C=0$ $\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 y-1}{\sqrt{3}}\right)=\frac{x^{3}}{3}+\frac{x^{2}}{2}+x$ put $\mathrm{x}=1$ $\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \mathrm{y}-1}{\sqrt{3}}\right)=\frac{1}{3}+\frac{1}{2}+1=\frac{2+3+6}{6}=\frac{11}{6}$ $\tan ^{-1}\left(\frac{2 y-1}{\sqrt{3}}\right)=\frac{11 \sqrt{3}}{12}$ $\frac{2 y-1}{\sqrt{3}}=\tan \left(\frac{11 \sqrt{3}}{12}\right)$ $\Rightarrow 2 y(1)-1=\sqrt{3} \tan \left(\frac{11 \sqrt{3}}{12}\right)$