AktiraAktira JEE
Subscribe
MathsJEE Main 4 Apr 2026, Shift 2·Q20·MCQ·Moderate

Definite Integration · Inverse trigonometric integrals

The integral ∫01cot⁡−1(1+x+x2)dx\displaystyle\int_0^1\cot^{-1}\left(1 + x + x^2\right)dx is equal to:
  1. (1)2tan⁡−12+12log⁡e(54)+π22\tan^{-1}2 + \dfrac12\log_e\left(\dfrac54\right) + \dfrac{\pi}{2}
  2. (2)2tan⁡−12+12log⁡e(54)−π22\tan^{-1}2 + \dfrac12\log_e\left(\dfrac54\right) - \dfrac{\pi}{2}
  3. (3)2tan⁡−12−12log⁡e(54)+π22\tan^{-1}2 - \dfrac12\log_e\left(\dfrac54\right) + \dfrac{\pi}{2}
  4. (4)2tan⁡−12−12log⁡e(54)−π22\tan^{-1}2 - \dfrac12\log_e\left(\dfrac54\right) - \dfrac{\pi}{2}

Video solution coming soon

We're animating every PYQ step by step. Subscribe to get them first.

Subscribe

See the full 4 Apr 2026, Shift 2 paper