Definite IntegrationmediumPYQ · JEE Main · 4 Apr 2026 · Shift 2 (Afternoon)Free

Integral of cot⁻¹(1+x+x²) from 0 to 1 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
The integral ∫01cot⁡−1(1+x+x2)dx\displaystyle\int_0^{1}\cot^{-1}\left(1+x+x^2\right)dx is equal to
A2tan⁡−12+12log⁡e(54)+π22\tan^{-1}2+\dfrac12\log_e\left(\dfrac54\right)+\dfrac{\pi}{2}
B2tan⁡−12+12log⁡e(54)−π22\tan^{-1}2+\dfrac12\log_e\left(\dfrac54\right)-\dfrac{\pi}{2}
C2tan⁡−12−12log⁡e(54)+π22\tan^{-1}2-\dfrac12\log_e\left(\dfrac54\right)+\dfrac{\pi}{2}
D2tan⁡−12−12log⁡e(54)−π22\tan^{-1}2-\dfrac12\log_e\left(\dfrac54\right)-\dfrac{\pi}{2}correct
Solution
Step 1: 1+x+x2=1+x(x+1)1+x+x^2=1+x(x+1):
cot⁡−1(1+x+x2)=tan⁡−111+x(x+1)=tan⁡−1(x+1)−x1+(x+1)x=tan⁡−1(x+1)−tan⁡−1x.\cot^{-1}(1+x+x^2)=\tan^{-1}\frac{1}{1+x(x+1)}=\tan^{-1}\frac{(x+1)-x}{1+(x+1)x}=\tan^{-1}(x+1)-\tan^{-1}x.
 Step 2:
I=∫01tan⁡−1(x+1) dx−∫01tan⁡−1x dx.I=\int_0^1\tan^{-1}(x+1)\,dx-\int_0^1\tan^{-1}x\,dx.
 Step 3: With ∫tan⁡−1u du=utan⁡−1u−12ln⁡(1+u2)\int\tan^{-1}u\,du=u\tan^{-1}u-\tfrac12\ln(1+u^2), substitute u=x+1u=x+1 in the first (du=dxdu=dx, limits 1→21\to2):
∫01tan⁡−1(x+1) dx=∫12tan⁡−1u du=[utan⁡−1u−12ln⁡(1+u2)]12\int_0^1\tan^{-1}(x+1)\,dx=\int_1^2\tan^{-1}u\,du=\Big[u\tan^{-1}u-\tfrac12\ln(1+u^2)\Big]_1^2
=(2tan⁡−12−12ln⁡5)−(π4−12ln⁡2)=2tan⁡−12−π4−12ln⁡52.=\left(2\tan^{-1}2-\tfrac12\ln5\right)-\left(\tfrac\pi4-\tfrac12\ln2\right)=2\tan^{-1}2-\frac\pi4-\frac12\ln\frac52.
 Step 4:
∫01tan⁡−1x dx=[xtan⁡−1x−12ln⁡(1+x2)]01=π4−12ln⁡2.\int_0^1\tan^{-1}x\,dx=\Big[x\tan^{-1}x-\tfrac12\ln(1+x^2)\Big]_0^1=\frac\pi4-\frac12\ln2.
 ∴\therefore
I=(2tan⁡−12−π4−12ln⁡52)−(π4−12ln⁡2)=2tan⁡−12−12ln⁡54−π2.I=\left(2\tan^{-1}2-\frac\pi4-\frac12\ln\frac52\right)-\left(\frac\pi4-\frac12\ln2\right)=2\tan^{-1}2-\frac12\ln\frac54-\frac\pi2.
 Correct answer: (4)
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