If 0 1 4 cot - 1 ( 1 - 2 x + 4 x 2 ) ? d x = a tan - 1 ( 2 ) - b log e ( 5 ) , where a , b , t h e n ( 2 a + b ) is equal to _______. [2026]

Question

If $\int_{0}^{1} 4 \cot^{- 1} (1 - 2 x + 4 x^{2}) d x = a \tan^{- 1} (2) - b \log_{e} (5),$ where $a, b \in ℕ, then (2 a + b)$ is equal to _______. [2026]

Smart Achievers · Accepted Answer

(9)

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

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

Applying king

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

From (1) & (2)

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

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

Applying King

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

$= \int_{0}^{1} (π - \cot^{- 1} (2 x)) d x - \int_{0}^{1} \cot^{- 1} (2 x) d x$

$= \int_{0}^{1} (π - 2 \cot^{- 1} (2 x)) d x$

$= π - 2 \int_{0}^{1} \cot^{- 1} (2 x) d x$

By parts

$= π - 2 [(x \cot^{- 1} {(2 x))}_{0}^{1} + \int_{0}^{1} \frac{2 x}{1 + 4 x^{2}} d x]$

Let $1 + 4 x^{2} = t$

$8 x d x = d t$

$= π - 2 [\cot^{- 1} 2 + \frac{1}{4} \int_{1}^{5} \frac{d t}{t}]$

$= π - 2 \cot^{- 1} 2 - \frac{1}{2} l n 5$

$2 I = 2 \tan^{- 1} 2 - \frac{1}{2} l n 5$

$\Rightarrow 4 I = 4 \tan^{- 1} 2 - l n 5$

$∴$ $2 a + b = 8 + 1 = 9$

Solution

Q.
If $\int_{0}^{1} 4 \cot^{- 1} (1 - 2 x + 4 x^{2}) d x = a \tan^{- 1} (2) - b \log_{e} (5),$ where $a, b \in ℕ, then (2 a + b)$ is equal to _______. [2026]

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Solution

Q. If ∫014cot-1(1-2x+4x2)dx=a tan-1(2)-bloge(5), where a,b∈ℕ, then (2a+b) is equal to _______. [2026]

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Q.
If $\int_{0}^{1} 4 \cot^{- 1} (1 - 2 x + 4 x^{2}) d x = a \tan^{- 1} (2) - b \log_{e} (5),$ where $a, b \in ℕ, then (2 a + b)$ is equal to _______. [2026]