求∫dx/(x^2+x+1)^2的不定积分

∫1/(x²+x+1)² dx
= ∫1/[(x+1/2)²+3/4]² dx
令x+1/2=√3/2*tanθ,dx=√3/2*sec²θ dθ
sinθ=(x+1/2)/√(x²+x+1),cosθ=(√3/2)/√(x²+x+1)
原式= (√3/2)∫sec²θ/(3/4*sec²θ)² dθ
= (√3/2)(16/9)∫sec²θ/sec⁴θ dθ
= 8/(3√3)*∫cos²θ dθ
= 4/(3√3)*∫(1+cos2θ) dθ
= 4/(3√3)*(θ+1/2*sin2θ) + C
= 4/(3√3)*arctan[(2x+1)/√3] + (2x+1)/[3(x²+x+1)] + C