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Integral of inverse cot of a quadratic from 0 to 1, split into a difference of inverse tans

The same question, asked 2 times: 21 Jan 2026 S2*, 4 Apr 2026 S2. Every wording below, oldest first.

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  1. Question 1

    21 Jan 2026 S2

    72
    If ∫014cot⁡−1(1−2x+4x2)dx=atan⁡−1(2)−bloge(5)\int_{0}^{1} 4\cot^{- 1}\left( 1 - 2x + 4x^{2} \right)dx = a\tan^{- 1}(2) - {blog}_{e}(5), where a,b∈Na,b \in N, then (2a+b)(2a + b) is equal to ____\_\_\_\_ .
  2. Question 2

    4 Apr 2026 S2

    70
    The integral ∫01cot⁡−1(1+x+x2)dx\int_{0}^{1} \cot^{- 1}\left( 1 + x +x^{2} \right)dx is equal to :

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