Question

求解这一题

Answer 1

∫ dx/[x(x^2+x+1)^2]
let
1/[x(x^2+x+1)^2]≡ A/x + (B1x+B2)/(x^2+x+1) + (C1x+C2)/(x^2+x+1)^2
=>
1≡ A(x^2+x+1)^2 + (B1x+B2)x(x^2+x+1) + (C1x+C2)x

x=0, =>A=1

coef. of x^4
A+B1=0
B1=-1

coef. of x^3
2A+B1+B2 =0
2-1+B2=0
B2=-1

coef. of x
2A+B2 +C2 =0
2-1+C2=0
C2 =-1

x=1
1= 9A + 3(B1+B2) + (C1+C2)
1=9 +3(-1-1)+(-1+C2)
1=9-6-1-C2
C2 =1

1/[x(x^2+x+1)^2]
=1/x + (-x-1)/(x^2+x+1) + (-x+1)/(x^2+x+1)^2
=1/x - (1/2)(2x+1)/(x^2+x+1) -(1/2) [1/(x^2+x+1)]
- (1/2)(2x+1)/(x^2+x+1)^2 + (3/2)[1/(x^2+x+1)^2]

∫ dx/[x(x^2+x+1)^2]
=ln|x| - (1/2)ln|(x^2+x+1)| -(1/2)∫ dx/(x^2+x+1) +(1/2)/(x^2+x+1)
+(3/2)∫ dx/(x^2+x+1)^2
=ln|x| - (1/2)ln|(x^2+x+1)| -(√3/3)arctan[(2x+1)/√3]
+(1/2)/(x^2+x+1) +(2√3/3)[ arctan[(2x+1)/√3] +(√3/4)(2x+1)/(x^2+x+1) ]+C

consider
x^2+x+1 = (x+1/2)^2 + 3/4
let
x+1/2 = (√3/2) tanu
dx =(√3/2)(secu)^2 .du

∫ dx/(x^2+x+1)
=∫ (√3/2)(secu)^2 .du /((3/4)(secu)^2)
=(2√3/3)∫ du
=(2√3/3)u + C'
=(2√3/3)arctan[(2x+1)/√3] + C'

∫ 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) + C''
=(4√3/9)[ arctan[(2x+1)/√3] +(√3/4)(2x+1)/(x^2+x+1) ] + C''

追问

好复杂的样子。