JEE Main 2026 — 21 January, Evening Shift — Question 22

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

Applying king $\begin{gathered} I=\int_{0}^{1}\left(-\cot ^{-1}(2 x-1)+\cot ^{-1}(2 x-2)\right) d x \end{gathered}$

From(1) & (2)

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

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

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

$=\int_{0}^{1}\left(\pi-\cot ^{-1}(2 \mathrm{x})\right) \mathrm{dx}-\int_{0}^{1} \cot ^{-1}(2 \mathrm{x}) \mathrm{dx}$ $=\int_{0}^{1}\left(\pi-2 \cot ^{-1}(2 x)\right) d x$ $=\pi-2 \int_{0}^{1}\left(\cot ^{-1} 2 \mathrm{x}\right) \cdot 1 \mathrm{dx}$

By parts $=\pi-2\left[\left(\mathrm{x}^{-1} 2 \mathrm{x}\right)_{0}^{1}+\int_{0}^{1} \frac{2 \mathrm{x}}{1+4 \mathrm{x}^{2}} \mathrm{dx}\right]$

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

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

$=\pi-2\left[\cot ^{-1} 2+\frac{1}{4} \int_{1}^{5} \frac{\mathrm{dt}}{\mathrm{t}}\right]$

$=\pi-2 \cot ^{-1} 2-\frac{1}{2} \ell \mathrm{n} 5$

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

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

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