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JEE Main 20264 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
4 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2026 · 4 April · Evening Shift · Mathematics · Definite Integration

The integral ∫_ 0 ^ 1 ^ -1 (1+x+x^2)dx is equal to:

The integral ∫01cot⁡−1(1+x+x2)dx\int_{0}^{1}\cot^{-1}(1+x+x^2)dx is equal to:

Options

  1. A2\tan^{-1}2+\dfrac{1}{2}\log_e(54)+π2\left(\dfrac{5}{4}\right)+\dfrac{\pi}{2}
  2. B2\tan^{-1}2+\dfrac{1}{2}\log_e(54)−π2\left(\dfrac{5}{4}\right)-\dfrac{\pi}{2}
  3. C2\tan^{-1}2-\dfrac{1}{2}\log_e(54)+π2\left(\dfrac{5}{4}\right)+\dfrac{\pi}{2}
  4. D2\tan^{-1}2-\dfrac{1}{2}\log_e(54)−π2\left(\dfrac{5}{4}\right)-\dfrac{\pi}{2}

Correct answer

D. 2\tan^{-1}2-\dfrac{1}{2}\log_e(54)−π2\left(\dfrac{5}{4}\right)-\dfrac{\pi}{2}

Step-by-step solution

I=∫01cot⁡−1(1+x+x2)dxI = \int_{0}^{1}\cot^{-1}(1+x+x^2)dx

I=∫01tan⁡−1(11+x+x2)dxI = \int_{0}^{1}\tan^{-1}\left(\dfrac{1}{1+x+x^2}\right)dx

I=∫01tan⁡−1((x+1)−x1+(x+1)x)dxI = \int_{0}^{1}\tan^{-1}\left(\dfrac{(x+1)-x}{1+(x+1)x}\right)dx

I=∫01(tan⁡−1(x+1)−tan⁡−1x)dxI = \int_{0}^{1}(\tan^{-1}(x+1)-\tan^{-1}x)dx

I=∫01tan⁡−1(x+1)dx−∫01tan⁡−1xdxI = \int_{0}^{1}\tan^{-1}(x+1)dx - \int_{0}^{1}\tan^{-1}x dx

Substituting x+1=tx+1 = t in the first integral:

I=∫12tan⁡−1xdx−∫01tan⁡−1xdxI = \int_{1}^{2}\tan^{-1}x dx - \int_{0}^{1}\tan^{-1}x dx

Using integration by parts, ∫tan⁡−1xdx=xtan⁡−1x−12log⁡e(1+x2)\int \tan^{-1}x dx = x\tan^{-1}x - \dfrac{1}{2}\log_e(1+x^2)

I=[xtan⁡−1x−12log⁡e(1+x2)]12−[xtan⁡−1x−12log⁡e(1+x2)]01I = \left[x\tan^{-1}x - \dfrac{1}{2}\log_e(1+x^2)\right]_{1}^{2} - \left[x\tan^{-1}x - \dfrac{1}{2}\log_e(1+x^2)\right]_{0}^{1}

I=(2tan⁡−12−12log⁡e5−(π4−12log⁡e2))−(π4−12log⁡e2−0)I = \left(2\tan^{-1}2 - \dfrac{1}{2}\log_e 5 - \left(\dfrac{\pi}{4} - \dfrac{1}{2}\log_e 2\right)\right) - \left(\dfrac{\pi}{4} - \dfrac{1}{2}\log_e 2 - 0\right)

I=2tan⁡−12−12log⁡e5−π2+log⁡e2I = 2\tan^{-1}2 - \dfrac{1}{2}\log_e 5 - \dfrac{\pi}{2} + \log_e 2

I=2tan⁡−12−12log⁡e5+12log⁡e4−π2I = 2\tan^{-1}2 - \dfrac{1}{2}\log_e 5 + \dfrac{1}{2}\log_e 4 - \dfrac{\pi}{2}

I=2tan⁡−12−12log⁡e(54)−π2I = 2\tan^{-1}2 - \dfrac{1}{2}\log_e\left(\dfrac{5}{4}\right) - \dfrac{\pi}{2}

Answer: 2tan⁡−12−12log⁡e(54)−π22\tan^{-1}2-\dfrac{1}{2}\log_e\left(\dfrac{5}{4}\right)-\dfrac{\pi}{2}
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