JEE Main 2026 · 4 Apr · Shift 2 · Mathematics Q20

← JEE Main 2026 · 4 Apr · Shift 2

MathematicsSingle correct+4 / -1Question 20 of 75

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

2\tan^{-1}2+\dfrac12\log_e\left(\dfrac54\right)+\dfrac\pi2

2\tan^{-1}2+\dfrac12\log_e\left(\dfrac54\right)-\dfrac\pi2

2\tan^{-1}2-\dfrac12\log_e\left(\dfrac54\right)+\dfrac\pi2

2\tan^{-1}2-\dfrac12\log_e\left(\dfrac54\right)-\dfrac\pi2

Answer and worked solution

Official answer: D

Split the integrand. $\cot^{-1}(1+x+x^2)=\tan^{-1}(x+1)-\tan^{-1}x$ .

Integrate. With $F(t)=\int_0^t\tan^{-1}=t\tan^{-1}t-\frac12\ln(1+t^2)$ , the integral is $F(2)-2F(1)$ :
$2\tan^{-1}2-\tfrac12\ln5-\tfrac\pi2+\ln2=2\tan^{-1}2-\tfrac12\ln\tfrac54-\tfrac\pi2.$

(D)

Topics:Properties of definite integrals Inverse trigonometric functions

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JEE Main 2026 · 4 Apr · Shift 2, Mathematics question 20