Q93 / 150 · MHT CET 2026 (16 April Shift 2) · Mathematics

The value of integral $\int_0^1\cot^{-1}(1 + x^2 - x)\,dx$ is...

  • Correct answer
    A
    $\dfrac{\pi}{2} - \log 2$
  • B
    $\pi - \log 2$
  • C
    $\dfrac{\pi}{4} - \log 2$
  • D
    $2\pi - \log 2$

Solution

$I = \int_0^1 \cot^{-1}(1 + x^2 - x) \, dx$

$I = \int_0^1 \tan^{-1}\left(\dfrac{1}{1 + x(x - 1)}\right) \, dx$

$I = \int_0^1 \tan^{-1}\left(\dfrac{x - (x - 1)}{1 + x(x - 1)}\right) \, dx$

$I = \int_0^1 (\tan^{-1}x - \tan^{-1}(x - 1)) \, dx$

Using the property $\int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx$ on the second term:

$\int_0^1 \tan^{-1}(x - 1) \, dx = \int_0^1 \tan^{-1}(1 - x - 1) \, dx = \int_0^1 \tan^{-1}(-x) \, dx = -\int_0^1 \tan^{-1}x \, dx$

$I = \int_0^1 \tan^{-1}x \, dx - \left(-\int_0^1 \tan^{-1}x \, dx\right) = 2\int_0^1 \tan^{-1}x \, dx$

Using integration by parts:

$I = 2 \left[ x \tan^{-1}x \right]_0^1 - 2 \int_0^1 \dfrac{x}{1 + x^2} \, dx$

$I = 2 \left( 1 \cdot \dfrac{\pi}{4} - 0 \right) - \left[ \log(1 + x^2) \right]_0^1$

$I = \dfrac{\pi}{2} - \log 2$

Answer: $\dfrac{\pi}{2} - \log 2$

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

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