不定积分1/x^3一1dx

Question

不定积分 1/x^3一1dx

Answer 1

∫ dx/(x^3-1)
let
1/(x^3-1) ≡ A/(x-1)+(Bx+C)/(x^2+x+1)
=>
1≡ A(x^2+x+1)+(Bx+C)(x-1)
x=1=> A=1/3
coef.of x^2
A+B=0
B= -1/3
coef.of constant
A-C = 1
C= -2/3
ie
1/(x^3-1) ≡(1/3) [1/(x-1)- (x+2) /(x^2+x+1)]
∫ dx/(x^3-1)
= (1/3)∫ [1/(x-1)-(x+2)/(x^2+x+1)]dx
=(1/3)[ ln|x-1| - (1/2)∫ (2x+1)/(x^2+x+1)]dx - (3/2)∫ dx/(x^2+x+1) ]
= (1/3){ ln|x-1| - (1/2)ln|x^2+x+1| + (3/2)∫ dx/(x^2+x+1) }
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/(x^2+x+1)
=(2√3/3)∫ dy
=(2√3/3)y + C'
=(2√3/3)arctan[(2x+1)/√3] + C'
∫ dx/(x^3-1)
= (1/3) { ln|x-1| - (1/2)ln|x^2+x+1| + (3/2)∫ dx/(x^2+x+1) }
= (1/3) { ln|x-1| - (1/2)ln|x^2+x+1| + √3arctan[(2x+1)/√3] }+ C