∫(-x^2-2)dx/(1+x+x^2)^2

题目

∫(-x^2-2)dx/(1+x+x^2)^2

答案

consider
-x^2-2 = -(x^2+x+1) +x-1
= -(x^2+x+1) +(1/2)(2x+1) - 3/2
∫(-x^2-2)/(1+x+x^2)^2 dx
=-∫dx/(1+x+x^2) +(1/2)∫(2x+1)/(1+x+x^2)^2 dx - (3/2)∫dx/(1+x+x^2)^2
=-∫dx/(1+x+x^2) -(1/2)[1/(1+x+x^2)] - (3/2)∫dx/(1+x+x^2)^2
consider
x^2+x+1 = (x+1/2)^2 + 3/4
let
x+ 1/2 = (√3/2) tany
dx =(√3/2) (secy)^2 dy

∫dx/(1+x+x^2) =(2√3/3)∫ dy
=(2√3/3)y
=(2√3/3)arctan[(2x+1)/√3]
∫dx/(1+x+x^2)^2 = (8√3/9)∫ (cosy)^2 dy
= (4√3/9)∫ (1+cos2y) dy
= (4√3/9) (y+sin(2y)/ 2)
= (4√3/9) { arctan[(2x+1)/√3]+√3(2x+1)/ [4(x^2+x+1)] }
∫(-x^2-2)/(1+x+x^2)^2 dx
=-∫dx/(1+x+x^2) -(1/2)[1/(1+x+x^2)] - (3/2)∫dx/(1+x+x^2)^2
=-(2√3/3)arctan[(2x+1)/√3] -(1/2)[1/(1+x+x^2)]
-(2√3/3){ arctan[(2x+1)/√3]+√3(2x+1)/[4(x^2+x+1)] } + C