Q45 / 75 · JEE Main 2026 (04 April Shift 2) · Mathematics

The integral $\int_{0}^{1}\cot^{-1}(1+x+x^2)dx$ is equal to:

  • A
    $2\tan^{-1}2+\dfrac{1}{2}\log_e\left(\dfrac{5}{4}\right)+\dfrac{\pi}{2}$
  • B
    $2\tan^{-1}2+\dfrac{1}{2}\log_e\left(\dfrac{5}{4}\right)-\dfrac{\pi}{2}$
  • C
    $2\tan^{-1}2-\dfrac{1}{2}\log_e\left(\dfrac{5}{4}\right)+\dfrac{\pi}{2}$
  • Correct answer
    D
    $2\tan^{-1}2-\dfrac{1}{2}\log_e\left(\dfrac{5}{4}\right)-\dfrac{\pi}{2}$

Solution

$I = \int_{0}^{1}\cot^{-1}(1+x+x^2)dx$

$I = \int_{0}^{1}\tan^{-1}\left(\dfrac{1}{1+x+x^2}\right)dx$

$I = \int_{0}^{1}\tan^{-1}\left(\dfrac{(x+1)-x}{1+(x+1)x}\right)dx$

$I = \int_{0}^{1}(\tan^{-1}(x+1)-\tan^{-1}x)dx$

$I = \int_{0}^{1}\tan^{-1}(x+1)dx - \int_{0}^{1}\tan^{-1}x dx$

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

$I = \int_{1}^{2}\tan^{-1}x dx - \int_{0}^{1}\tan^{-1}x dx$

Using integration by parts, $\int \tan^{-1}x dx = x\tan^{-1}x - \dfrac{1}{2}\log_e(1+x^2)$

$I = \left[x\tan^{-1}x - \dfrac{1}{2}\log_e(1+x^2)\right]_{1}^{2} - \left[x\tan^{-1}x - \dfrac{1}{2}\log_e(1+x^2)\right]_{0}^{1}$

$I = \left(2\tan^{-1}2 - \dfrac{1}{2}\log_e 5 - \left(\dfrac{\pi}{4} - \dfrac{1}{2}\log_e 2\right)\right) - \left(\dfrac{\pi}{4} - \dfrac{1}{2}\log_e 2 - 0\right)$

$I = 2\tan^{-1}2 - \dfrac{1}{2}\log_e 5 - \dfrac{\pi}{2} + \log_e 2$

$I = 2\tan^{-1}2 - \dfrac{1}{2}\log_e 5 + \dfrac{1}{2}\log_e 4 - \dfrac{\pi}{2}$

$I = 2\tan^{-1}2 - \dfrac{1}{2}\log_e\left(\dfrac{5}{4}\right) - \dfrac{\pi}{2}$

Answer: $2\tan^{-1}2-\dfrac{1}{2}\log_e\left(\dfrac{5}{4}\right)-\dfrac{\pi}{2}$

Hence, the correct answer is option: 'D'.

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