JEE Main 2026 — Mathematics Calculus

Let $\displaystyle I=\int_0^1 \cot ^{-1}\left(1-2 x+4 x^2\right) d x$I=∫01​cot−1(1−2x+4x2)dx

$$I=\int_0^1\left(\cot ^{-1}(2 x-1)-\cot ^{-1}(2 x)\right) d x$$I=∫01​(cot−1(2x−1)−cot−1(2x))dx

Applying king

$$I=\int_0^1\left(-\cot ^{-1}(2 x-1)+\cot ^{-1}(2 x-2)\right) d x$$I=∫01​(−cot−1(2x−1)+cot−1(2x−2))dx

From (1) and (2)

$$2 I=\int_0^1\left(\cot ^{-1}(2 x-2)-\cot ^{-1}(2 x)\right) d x$$2I=∫01​(cot−1(2x−2)−cot−1(2x))dx

$$=\int_0^1 \cot ^{-1}(2 x-2) d x-\int_0^1 \cot ^{-1}(2 x) d x$$=∫01​cot−1(2x−2)dx−∫01​cot−1(2x)dx

Applying King

$$=\int_0^1 \cot ^{-1}(-2 x) d x-\int_0^1 \cot ^{-1}(2 x) d x$$=∫01​cot−1(−2x)dx−∫01​cot−1(2x)dx

$$=\int_0^1\left(\pi-\cot ^{-1}(2 x)\right) d x-\int_0^1 \cot ^{-1}(2 x) d x$$=∫01​(π−cot−1(2x))dx−∫01​cot−1(2x)dx

$$=\int_0^1\left(\pi-2 \cot ^{-1}(2 x)\right) d x$$=∫01​(π−2cot−1(2x))dx

$$=\pi-2 \int_0^1\left(\cot ^{-1} 2 x\right) \cdot 1 d x$$=π−2∫01​(cot−12x)⋅1dx

By parts

$$=\pi-2\left[\left(x \cot ^{-1} 2 x\right)_0^1+\int_0^1 \frac{2 x}{1+4 x^2} d x\right]$$=π−2[(xcot−12x)01​+∫01​1+4x22x​dx]

Let $1+4 \mathrm{x}^2=\mathrm{t}$1+4x2=t

$$8 \mathrm{xdx}=\mathrm{dt}$$8xdx=dt

$$=\pi-2\left[\cot ^{-1} 2+\frac{1}{4} \int_1^5 \frac{\mathrm{dt}}{\mathrm{t}}\right]$$=π−2[cot−12+41​∫15​tdt​]

$$=\pi-2 \cot ^{-1} 2-\frac{1}{2} \ell \mathrm{n} 5$$=π−2cot−12−21​ℓn5

$$2 \mathrm{I}=2 \tan ^{-1} 2-\frac{1}{2} \ell \mathrm{n} 5$$2I=2tan−12−21​ℓn5

$$\Rightarrow 4 \mathrm{I}=4 \tan ^{-1} 2-\ell \mathrm{n} 5$$⇒4I=4tan−12−ℓn5

$$\therefore 2 \mathrm{a}+\mathrm{b}=8+1=9$$∴2a+b=8+1=9