Question

高数求不定积分，谢谢哈

Answer 1

x^3+1 = (x+1)(x^2-x +1)
let
1/(x^3+1) ≡ A/(x+1) + (B1x+B2)/(x^2-x +1)
=>
1 ≡ A(x^2-x +1) + (B1x+B2)(x+1)

x=-1 =>A =1/3

coef. of constant
A+B2 = 1
B2 = 2/3

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

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| - (1/6)∫[(2x-1)/(x^2-x +1) ] dx +(1/2)∫ dx/(x^2-x +1)
=(1/3)ln|x+1| - (1/6)ln|x^2-x +1| +(1/2)∫ 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| - (1/6)ln|x^2-x +1| +(1/2)∫ dx/(x^2-x +1)
=(1/3)ln|x+1| - (1/6)ln|x^2-x +1| +(√3/3)arctan[(2x-1)/√3] + C