Let $\mathrm{I}=\int_{0}^{1} \cot ^{-1}\left(1-2 \mathrm{x}+4 \mathrm{x}^{2}\right) \mathrm{dx}\\$ I=\int_{0}^{1}\left(\cot ^{-1}(2 x-1)-\cot ^{-1}(2 x)\right) d x \tag{1}\\ Applying king \\ I=\int_{0}^{1}\left(-\cot ^{-1}(2 x-1)+\cot ^{-1}(2 x-2)\right) d x \tag....{(2)}\\ From (1) & (2) $\\2 \mathrm{I}=\int_{0}^{1}\left(\cot ^{-1}(2 \mathrm{x}-2)-\cot ^{-1}(2 \mathrm{x})\right) \mathrm{dx}\\$ $=\int_{0}^{1} \cot ^{-1}(2 \mathrm{x}-2) \mathrm{dx}-\int_{0}^{1} \cot ^{-1}(2 \mathrm{x}) \mathrm{dx}\\$ Applying King $\\=\int_{0}^{1} \cot ^{-1}(-2 x) d x-\int_{0}^{1} \cot ^{-1}(2 x) d x\\$ $=\int_{0}^{1}\left(\pi-\cot ^{-1}(2 \mathrm{x})\right) \mathrm{dx}-\int_{0}^{1} \cot ^{-1}(2 \mathrm{x}) \mathrm{dx}\\$ $=\int_{0}^{1}\left(\pi-2 \cot ^{-1}(2 x)\right) d x\\$ $=\pi-2 \int_{0}^{1}\left(\cot ^{-1} 2 \mathrm{x}\right) \cdot 1 \mathrm{dx}\\$ By parts $\\ \pi - 2 \left[ \left( x \cot^{-1} 2x \right)_0^1 + \int_0^1 \frac{2x}{1 + 4x^2} dx \right] \\$ $\\$Let $1+4 \mathrm{x}^{2}=\mathrm{t}\\$ $8 \mathrm{xdx}=\mathrm{dt}\\$ $=\pi-2\left[\cot ^{-1} 2+\frac{1}{4} \int_{1}^{5} \frac{\mathrm{dt}}{\mathrm{t}}\right]\\$ $=\pi-2 \cot ^{-1} 2-\frac{1}{2} \ell \mathrm{n} 5\\$ $2 \mathrm{I}=2 \tan ^{-1} 2-\frac{1}{2} \ln 5\\$ $\Rightarrow 4 \mathrm{I}=4 \tan ^{-1} 2-\ell \mathrm{n} 5\\$ $\therefore 2 \mathrm{a}+\mathrm{b}=8+1=9$