The value of ∫ 0 1 cot - 1 1 - x + x 2 d x is

${\cot }^{-1}\left(1-x+x^2\right)={\tan }^{-1}\left(\frac{1}{1-x+x^2}\right)$

$\Rightarrow {\cot }^{-1}\left(1-x+x^2\right)={\tan }^{-1}\left(\frac{1}{1-x\left(1-x\right)}\right)$

$\Rightarrow {\cot }^{-1}\left(1-x+x^2\right)={\tan }^{-1}\left(\frac{x+\left(1-x\right)}{1-x\left(1-x\right)}\right)$

$\Rightarrow {\cot }^{-1}\left(1-x+x^2\right)={\tan }^{-1}x+{\tan }^{-1}\left(1-x\right)$

$\therefore {\int }_0^1{\cot }^{-1}\left(1-x+x^2\right)dx={\int }_0^1{\tan }^{-1}xdx+{\int }_0^1{\tan }^{-1}\left(1-x\right)dx$

$\Rightarrow {\int }_0^1{\cot }^{-1}\left(1-x+x^2\right)dx={\int }_0^1{\tan }^{-1}xdx+{\int }_0^1{\tan }^{-1}xdx$   $\left[\because {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx\right]$

$\Rightarrow {\int }_0^1{\cot }^{-1}\left(1-x+x^2\right)dx=2{\int }_0^1{\tan }^{-1}xdx$

On evaluating by integration by parts, we have

$\Rightarrow {\int }_0^1{\cot }^{-1}\left(1-x+x^2\right)dx=2\left\{{\left[{\tan }^{-1}x·x\right]}_0^1-{\int }_0^1\frac{x}{1+x^2}dx\right\}$

$\Rightarrow {\int }_0^1{\cot }^{-1}\left(1-x+x^2\right)dx=2\left\{\frac{\pi }{4}-{\left[\frac{1}{2}\ln \left(1+x^2\right)\right]}_0^1\right\}$

$\Rightarrow {\int }_0^1{\cot }^{-1}\left(1-x+x^2\right)dx=2\left[\frac{\pi }{4}-\frac{1}{2}{\log }_{e}2\right]=\frac{\pi }{2}-{\log }_{e}2$

Hence, ${\int }_0^1{\cot }^{-1}\left(1-x+x^2\right)dx=\frac{\pi }{2}-{\log }_{e}2$

So, option $\left(b\right)$ is correct.