Notice $1-2x+4x^2=1+2x(2x-1)$, which suggests rewriting ${\cot }^{-1}$ using the identity ${\cot }^{-1}\left(\frac{1+AB}{B-A}\right)={\cot }^{-1}A-{\cot }^{-1}B$ (taking $A=2x-1$, $B=2x$).

So ${\cot }^{-1}(1-2x+4x^2)={\cot }^{-1}(2x-1)-{\cot }^{-1}(2x)$ (after checking the sign of $B-A=1>0$).

Let $I={\int }_0^1\left[{\cot }^{-1}(2x-1)-{\cot }^{-1}(2x)\right]dx$. Apply the king-property substitution $x\to 1-x$ on the same integral; combine to get $2I={\int }_0^1\left[{\cot }^{-1}(2x-2)-{\cot }^{-1}(2x)\right]dx$. Further manipulation with King once more reduces the integrand to $\pi -2{\cot }^{-1}(2x)$, yielding $2I=\pi -2{\int }_0^1{\cot }^{-1}(2x)dx$.

Now ${\int }_0^1{\cot }^{-1}(2x)dx$ by parts (taking $u={\cot }^{-1}(2x)$, $dv=dx$): $=[x{\cot }^{-1}(2x){]}_0^1+{\int }_0^1\frac{2x}{1+4x^2}dx={\cot }^{-1}(2)+\frac{1}{4}\ln 5$.

So $2I=\pi -2{\cot }^{-1}(2)-\frac{1}{2}\ln 5=2{\tan }^{-1}(2)-\frac{1}{2}\ln 5$, giving $4I=4{\tan }^{-1}(2)-\ln 5$. Therefore the original integral ${\int }_0^{1}4{\cot }^{-1}(\dots )dx=4I=4{\tan }^{-1}(2)-\ln 5$.

Match: $a=4$, $b=1$, so $2a+b=9$.