If $\int_{0}^{1} 4 \cot ^{-1}\left(1-2 \mathrm{x}+4 \mathrm{x}^{2}\right) \mathrm{dx}=\mathrm{a} \tan ^{-1}(2)-\mathrm{b}\log_{\mathrm{e}}(5)$, where $\mathrm{a}, \mathrm{b} \in \mathbf{N}$, then $(2 \mathrm{a}+\mathrm{b})$ is equal to ____.

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Definite Integrals

\int_{0}^{1} 4 cot^{- 1} (1 - 2 x + 4 x^{2}) dx = a tan^{- 1} (2) - b lo g_{e} (5)

, where

a, b \in N

, then

(2 a + b)

is equal to ____.

Answer: 9

Since

4 x^{2} - 2 x + 1 = 1 + (2 x) (2 x - 1)

cot^{- 1} (1 - 2 x + 4 x^{2}) = tan^{- 1} \frac{1}{1 + ( 2 x ) ( 2 x - 1 )} = tan^{- 1} (2 x) - tan^{- 1} (2 x - 1)

. Hence

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

. The second integral is

0

, since

tan^{- 1} (2 x - 1)

is odd about

x = \frac{1}{2}

. For the first,

\int_{0}^{1} tan^{- 1} (2 x) d x = [x tan^{- 1} (2 x) - \frac{1}{4} ln (1 + 4 x^{2})]_{0}^{1} = tan^{- 1} 2 - \frac{1}{4} ln 5

. Multiplying by

4

\int_{0}^{1} 4 cot^{- 1} (1 - 2 x + 4 x^{2}) d x = 4 tan^{- 1} 2 - ln 5

, so

a = 4, b = 1

and

2 a + b = 9

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Integer answer