Question

这个积分怎么做

Answer 1

答：
令x/(x^3-1) ≡ A/(x-1) + (Bx+C)/(x^2+x+1)
=>
x ≡ 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 =0
C =1/3
ie
x/(x^3-1) ≡(1/3)[ 1/(x-1) + (-x+1)/(x^2+x+1)]
∫[x/(x^3-1)]dx
=(1/3)∫[ 1/(x-1) + (-x+1)/(x^2+x+1)]dx
=(1/3)ln|x-1| - (1/6)∫[( 2x +1)/(x^2+x+1)]dx + (2/3)∫ dx/(x^2+x+1)
=(1/3)ln|x-1| - (1/6)ln|x^2+x+1| + (2/3)∫ dx/(x^2+x+1)
consider
x^2+x+1 = (x+ 1/2) + 3/4
let
x+1/2 = (√3/2)tany
dx = (√3/2)(secy)^2 dy
∫ dx/(x^2+x+1)
=∫ dx/(x^2+x+1)
=(2√3/3)∫ dy
=(2√3/3)y + C'
=(2√3/3)arctan[(2x+1)/√3] + C'
∫[x/(x^3-1)]dx
=(1/3)ln|x-1| - (1/6)ln|x^2+x+1| + (2/3)∫ dx/(x^2+x+1)
=(1/3)ln|x-1| - (1/6)ln|x^2+x+1| + (4√3/9)arctan[(2x+1)/√3] + C