Evaluate: `int_0^1 cot^-1(1-x+x^2)dx`

Explanation

`int_0^1 cot^-1(1-x+x^2)dx`
`=int_0^1 tan^-1(1/(1-x+x^2))dx`
`=int_0^1 tan^-1(1/(1-x(1-x)))dx`
`=int_0^1 tan^-1((x+1-x)/(1-x(1-x)))dx`
`=int_0^1 [tan^-1 x+tan^-1 (1-x)]dx`
Let `1-x=t`
`=>-dx=dt`
when `x->0 to 1`
`=>t-> 1 to 0 `
`=int_0^1 tan^-1 xdx-int_0^1 tan^-1 dt`
`=2 int _0^1 tab^-1 xdx `
`=2(pi/4-(log e^2)/2)`
`=pi/4-log^2`
` int_0^1 tan^-1 x dx`
using integration by parts
`=[xtan^-1 x]_0^1-int_0^1 x(1/(1+x^2))dx`
`=pi/4-1/2[log_e(1+x^2)]_^1`
`=pi/4-(log2)/2`
`=pi/4-log^2`