JEE Main 2026 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 4 April 2026, Shift 2, Maths Q20

Question 20 of 75 in this shift, Maths question 20 of 25, Section A.

The integral

\int_0^1\cot^{-1}\left(1+x+x^2\right)dx

is equal to:

(1) $2\tan^{-1}2+\frac12\log_e\left(\frac54\right)+\frac\pi2$
(2) $2\tan^{-1}2+\frac12\log_e\left(\frac54\right)-\frac\pi2$
(3) $2\tan^{-1}2-\frac12\log_e\left(\frac54\right)+\frac\pi2$
(4) $2\tan^{-1}2-\frac12\log_e\left(\frac54\right)-\frac\pi2$ Official answer

Official answer

Option 4

NTA final key.

Chapter: Integrals
Topic: Inverse Trig Identities
Idea tested: Piecewise cot-1 to tan-1 Conversion

Same topic in other shifts

All Inverse Trig Identities questions

4 Apr 2026, Shift 1 · Q16If $y=\tan^{-1}\left(\frac{3\cos x-4\sin x}{4\cos x+3\sin x}\right)+2\tan^{-1}\left(\frac{x}{1+\sqrt{1-x^2}}\right)$ , then $\frac{dy}{dx}$ …MediumSingle correct
5 Apr 2026, Shift 1 · Q24If $\frac{\pi}{4} + \sum_{p=1}^{11} \tan^{-1}\left(\frac{2^{p-1}}{1 + 2^{2p-1}}\right) = \alpha$ , then $\tan\alpha$ is equal to ________.MediumNumerical value
6 Apr 2026, Shift 1 · Q13Let $0<\alpha<1$ , $\beta=\frac{1}{3\alpha}$ and $\tan^{-1}(1-\alpha)+\tan^{-1}(1-\beta)=\frac{\pi}{4}$ . Then $6(\alpha+\beta)$ is equal to:MediumSingle correct
6 Apr 2026, Shift 2 · Q13If $\sin\left(\tan^{-1}\left(x\sqrt{2}\right)\right)=\cot\left(\sin^{-1}\sqrt{1-x^2}\right)$ , $x\in(0,1)$ , then the value of $x$ is :MediumSingle correct
8 Apr 2026, Shift 2 · Q12Let $\alpha = 3\sin^{-1}\left(\frac{6}{11}\right)$ and $\beta = 3\cos^{-1}\left(\frac{4}{9}\right)$ , where inverse trigonometric functions…HardTwo statements
2 Apr 2025, Shift 2 · Q23If $y = \cos\left(\frac{\pi}{3} + \cos^{-1}\frac{x}{2}\right)$ , then $(x-y)^2 + 3y^2$ is equal to __________.MediumNumerical value

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.