Question

这种这么拆分，求它的积分

Answer 1

let
(x^3+4x^2+x)/[(x+2)^2.(x^2+x+1)]≡ A/(x+2)+B/(x+2)^2 + (Cx+D)/(x^2+x+1)

=>
x^3+4x^2+x≡ A(x+2)(x^2+x+1)+B(x^2+x+1) + (Cx+D)(x+2)^2
x=-2,
-8+16-2 =B(4-2+1)
3B=6
B=2
两边求导
x^3+4x^2+x≡ A(x+2)(x^2+x+1)+B(x^2+x+1) + (Cx+D)(x+2)^2
3x^2+8x+1≡ A(x^2+x+1)+A(x+2)(2x+1)+B(2x+1) + C(x+2)^2 +2(Cx+D)(x+2)
x=-2
12-16+1=A(4-2+1)+B(-4+1)
3A-3B=-3
3A-6=-3
A=1
x^3+4x^2+x≡ A(x+2)(x^2+x+1)+B(x^2+x+1) + (Cx+D)(x+2)^2
coef. of x^3
A+C=1
1+C=1
C=0
coef. of constant

2A+B+4D=0
2+2+4D=0
D=-1
ie
(x^3+4x^2+x)/[(x+2)^2.(x^2+x+1)]≡ 1/(x+2)+2/(x+2)^2 -1/(x^2+x+1)

x^2+x+1 = (x+1/2)^2 + 3/4
let
x+1/2 =(√3/2) tanu
dx =(√3/2)(secu)^2 du
∫(x^3+4x^2+x)/[(x+2)^2.(x^2+x+1)] dx
=∫ [1/(x+2)+2/(x+2)^2 -1/(x^2+x+1)] dx
=ln|x+2| -2/(x+2) -∫ dx/(x^2+x+1)
=ln|x+2| -2/(x+2) -∫ (√3/2)(secu)^2 du/[(3/4)(secu)^2]
=ln|x+2| -2/(x+2) -(2√3/3)u + C
=ln|x+2| -2/(x+2) -(2√3/3)arctan[(2x+1)/√3] + C