原式 I = - ∫[(x^2+x+1)-(x-1)]dx/(x^2+x+1)^2
= - ∫dx/(x^2+x+1) + ∫(x-1)dx/(x^2+x+1)^2 = I1 + I2 + C,
其中
I1 = - ∫d(x+1/2)/[(x+1/2)^2+3/4) = -(2/√3) arctan[(2x+1)/√3]；
令 x+1/2 = (√3/2)tanu, 则 x = (√3/2)tanu - 1/2，dx = (√3/2)(secu)^2du,
I2 = ∫(x-1)dx/[(x+1/2)^2+3/4)^2
= ∫[(√3/2)tanu-3/2] (√3/2)(secu)^2du/[(9/16)(secu)^4]
= (4/3)∫(tanu-√3)du/(secu)^2
= (4/3)∫[sinucosu-√3(cosu)^2]du
= - (1/3)cos2u - (2/√3) ∫ (1+cos2u)du
= - (1/3)cos2u - (2/√3) [(u+(1/2)sin2u)]
= - (1/3)cos2u - 2u/√3 - (1/√3) sin2u
= (-2/√3)arctan[(2x+1)/√3]
- (1/3)[1-(4/3)(x+1/2)^2]/[1+(4/3)(x+1/2)^2]
- (1/√3) (4/√3)(x+1/2)/[1+(4/3)(x+1/2)^2]
= (-2/√3)arctan[(2x+1)/√3] - (x^2+4x+1)/(x^2+x+1)
则 I = (-4/√3)arctan[(2x+1)/√3] - (x^2+4x+1)/(x^2+x+1) + C
= -(2/√3) arctan[(2x+1)/√3] + (1/2)∫d(x^2+x+1)/(x^2+x+1)^2 + (3/2)∫dx/(x^2+x+1)^2