A definite integral...

First we will write :

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

Our integral becomes $I=\int _{ 0 }^{ 1 }{ \cot ^{ -1 }{ { (x }^{ 2 }-x+1)dx } } =\int _{ 0 }^{ 1 }{ \tan ^{ -1 }{ x } dx } -\int _{ 0 }^{ 1 }{ \tan ^{ -1 }{ (x-1) } dx }$

Also in the integral $\int _{ 0 }^{ 1 }{ \tan ^{ -1 }{ (x-1) } dx }$ we put $t=1-x$ we get it as : $-\int _{ 0 }^{ 1 }{ \tan ^{ -1 }{ t } dt }$

$I=2\int _{ 0 }^{ 1 }{ \tan ^{ -1 }{ x } dx }$

Put $x=tan\theta$ to get $I=2\int _{ 0 }^{ \pi /4 }{ \theta { sec }^{ 2 }\theta d\theta }$

Apply integration by parts to get $I=2(\theta tan\theta { | }_{ 0 }^{ \pi /4 }-\int _{ 0 }^{ \pi /4 }{ tan\theta d\theta } )=2(\frac { \pi }{ 4 } -\frac { 1 }{ 2 } ln(2))=\boxed { \frac { \pi }{ 2 } -ln(2) }$