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

then ${\int }_0^1{\tan }^{-1}\left(1-x+x^2\right)dx$ is equal to 

We have,

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

$\begin{array}{l}\Rightarrow 2{\int }_0^1 {\tan }^{-1}xdx={\int }_0^1 \left\{\frac{\pi }{2}-{\tan }^{-1}\left(1-x+x^2\right)\right\}dx\\ \Rightarrow {\int }_0^1 {\tan }^{-1}\left(1-x+x^2\right)dx=\frac{\pi }{2}-2{\int }_0^1 {\tan }^{-1}xdx\\ \Rightarrow {\int }_0^1 {\tan }^{-1}\left(1-x+x^2\right)dx=\frac{\pi }{2}-2{\left[x{\tan }^{-1}x-\frac{1}{2}\log \left(1+x^2\right)\right]}_0^1\\ \Rightarrow {\int }_0^1 {\tan }^{-1}\left(1-x+x^2\right)dx=\frac{\pi }{2}-2\left(\frac{\pi }{4}-\frac{1}{2}\log 2\right)=\log 2\end{array}$