Integration

x^2 - x + 1 = (x - 1/2)^2 + (√3/2)^2

∫(0 to 1) 1/(x^2-x+1) dx

= ∫(0 to 1) (x - 1/2)^2 + (√3/2)^2 d(x - 1/2)

= 1/(√3/2) arctan (x - 1/2)/(√3/2) | [0,1]

= (2/√3)(π/6 + π/6)

= 2√3π/9

∫(0>1) dx/(x^2-x+1)

1/(x^2-x+1) = 1/[(x - 1/2)^2 + 3/4]
let x - 1/2 = y sqrt(3/4)
dx/dy = sqrt(3/4)
1/[(x - 1/2)^2 + 3/4] = 1/{3y^2/4 + 3/4}
1/[(x - 1/2)^2 + 3/4] = 4/{3[y^2 + 1]}
when x = 0 => y = -1/sqrt(3)
when x = 1 => y = 1/sqrt(3)

∫(0>1) dx/(x^2-x+1)
= ∫[-1/sqrt(3)>1/sqrt(3)] [sqrt(3/4)dy] [4/{3[y^2 + 1]}
= (4/3)sqrt(3/4) ∫[-1/sqrt(3)>1/sqrt(3)] dy /(y^2 + 1)
= 2/[sqrt(3)] ∫[-1/sqrt(3)>1/sqrt(3)] dy /(y^2 + 1)

let y = tan z
dy/dz = (sec z)^2
y^2 + 1 = (tan z)^2 + 1 = (sec z)^2
when y = -1/sqrt(3) => -pi/6
when y = 1/sqrt(3) => pi/6
2/[sqrt(3)] ∫[-1/sqrt(3)>1/sqrt(3)] dy /(y^2 + 1)
= 2/[sqrt(3)] ∫[-pi/6>pi/6] (sec z)^2 dz /[(sec z)^2]
= 2/[sqrt(3)] ∫[-pi/6>pi/6] dz
= 2/[sqrt(3)] [z] [-pi/6>pi/6]
= 2/[sqrt(3)] [pi/6 - (-pi/6)]
= 2/[sqrt(3)] [pi/3]
= 2pi/[3sqrt(3)]
= 2pi [sqrt(3)]/9