Aspire Faculty ID #17931 · Topic: JEE Main 2026 (21 January Evening Shift) · Just now
JEE Main 2026 (21 January Evening Shift)

If $\displaystyle \int_{0}^{1} 4\cot^{-1}(1 - 2x + 4x^2),dx = a\tan^{-1}(2) - b\log_e(5)$, where $a,b \in \mathbb{N}$, then $(2a + b)$ is equal to ______.

Solution

Let $I = \int_{0}^{1} \cot^{-1}(1 - 2x + 4x^2),dx$

$I = \int_{0}^{1} \left(\cot^{-1}(2x-1) - \cot^{-1}(2x)\right),dx \quad ...(1)$

Applying King

$I = \int_{0}^{1} \left(-\cot^{-1}(2x-1) + \cot^{-1}(2x-2)\right),dx \quad ...(2)$

From (1) & (2)

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

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

Applying King

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

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

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

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

By parts

$I = \pi - 2\left[x\cot^{-1}(2x)\right]{0}^{1} + \int{0}^{1} \frac{2x}{1+4x^2},dx$

Let $1 + 4x^2 = t$

$8x,dx = dt$

$I = \pi - 2\cot^{-1}(2) + \frac{1}{4}\int_{1}^{5} \frac{dt}{t}$

$= \pi - 2\cot^{-1}(2) + \frac{1}{4}\ln 5$

$\Rightarrow 2I = 2\pi - 4\cot^{-1}(2) + \frac{1}{2}\ln 5$

Given $\int_{0}^{1} 4\cot^{-1}(1 - 2x + 4x^2),dx = 4I$

$= 2\left[2\pi - 4\cot^{-1}(2) + \frac{1}{2}\ln 5\right]$

$= 4\pi - 8\cot^{-1}(2) + \ln 5$

$\Rightarrow 2a + b = 8 + 1 = 9$
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