JEE Main 2026 (Online) 4th April Evening ShiftQuestion 1
The integral $\int_{0}^{1}\cot^{-1}\left(1+x+x^{2}\right)dx$ is equal to:
Definite Integration
Solution
Concept. $\frac{1}{1+x+x^2}=\frac{(x+1)-x}{1+(x+1)x}$, so the integrand is a difference of two inverse tangents. Formulas used. (1) $\cot^{-1}y=\tan^{-1}\frac1y$ for $y>0$, and $\tan^{-1}a-\tan^{-1}b=\tan^{-1}\frac{a-b}{1+ab}$ ($ab>-1$)(2) $\int\tan^{-1}u\,du=u\tan^{-1}u-\frac12\ln(1+u^2)$Step 1 - Split. By (1), $$\cot^{-1}(1+x+x^2)=\tan^{-1}(x+1)-\tan^{-1}x.$$ Step 2 - Second part. By (2), $$\int_0^1\tan^{-1}x\,dx=\frac\pi4-\frac12\ln2.$$ Step 3 - First part. With $u=x+1$, by (2), $$\int_1^2\tan^{-1}u\,du=\Big(2\tan^{-1}2-\tfrac12\ln5\Big)-\Big(\tfrac\pi4-\tfrac12\ln2\Big).$$ Step 4 - Subtract. $$I=2\tan^{-1}2-\frac12\ln5+\ln2-\frac\pi2=2\tan^{-1}2-\frac12\ln\frac54-\frac\pi2.$$ Answer: (D)