JEE Main 2026 (4 Apr, Shift 2) Maths Q20: Integrals

Question

The integral $\int_0^1\cot^{-1}\left(1+x+x^2\right)dx$ is equal to:

Accepted Answer

Option 4: $2	an^{-1}2-\frac12\log_e\left(\frac54ight)-\frac\pi2$

Answer

$2	an^{-1}2+\frac12\log_e\left(\frac54ight)+\frac\pi2$

Answer

$2	an^{-1}2+\frac12\log_e\left(\frac54ight)-\frac\pi2$

Answer

$2	an^{-1}2-\frac12\log_e\left(\frac54ight)+\frac\pi2$

Answer

$2	an^{-1}2-\frac12\log_e\left(\frac54ight)-\frac\pi2$

JEE Main 4 April 2026, Shift 2, Maths Q20