The value of the Integral ${\int }_0^1 {\cot }^{-1}\left(1-x+x^2\right)dx$ is

$\begin{array}{l}{\cot }^{-1}\left(1-x+x^2\right)={\tan }^{-1}\frac{x-(x-1)}{1+x(x-1)}\\ ={\tan }^{-1}x-{\tan }^{-1}(x-1)\\ \therefore I={\int }_0^1 {\tan }^{-1}xdx-{\int }_0^1 {\tan }^{-1}(x-1)dx\end{array}$

Apply Prop. IV on 2nd and it becomes $+{\int }_0^1 {\tan }^{-1}xdx$

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