高数不定积分，求大神解答

(-x^2-2)/(x^2+x+1)^2
=-1/(x^2+x+1) + (x-1)/(x^2+x+1)^2
=-1/(x^2+x+1) + (1/2) [(2x+1)/(x^2+x+1)^2] - (3/2)[1/(x^2+x+1)^2]
consider
x^2+x+1 = (x+1/2)^2 + 3/4
let
x+1/2 = (√3/2)tanu
dx =(√3/2)(secu)^2 .du
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∫(-x^2-2)/(x^2+x+1)^2 dx
=-∫dx/(x^2+x+1) + (1/2)∫[(2x+1)/(x^2+x+1)^2] dx - (3/2)∫dx/(x^2+x+1)^2
=-(1/2)[1/(x^2+x+1)] -∫dx/(x^2+x+1) - (3/2)∫dx/(x^2+x+1)^2
=-(1/2)[1/(x^2+x+1)] -(2√3/3)arctan[(2x+1)/√3]
- (2√3/3){ arctan[(2x+1)/√3]+√3.(2x+1)/(x^2+x+1) } +C
=-(1/2)[1/(x^2+x+1)] -(4√3/3)arctan[(2x+1)/√3] - 2(2x+1)/(x^2+x+1) + C
consider
∫dx/(x^2+x+1)
=∫(√3/2)(secu)^2 .du/[(3/4)(secu)^2]
=(2√3/3)∫ du
=(2√3/3)u + C1
=(2√3/3)arctan[(2x+1)/√3] + C1
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∫dx/(x^2+x+1)^2
=∫(√3/2)(secu)^2 .du/ [(9/16) (secu)^4 ]
=(8√3/9) ∫ (cosu)^2 du
=(4√3/9) ∫ (1+cos2u) du
=(4√3/9)(u+(1/2)sin2u) +C2
=(4√3/9){ arctan[(2x+1)/√3]+√3.(2x+1)/(x^2+x+1) } +C2