Question

反常积分:1/(1 x^3)的原函数是什么？

Answer 1

1/(1+x^3) = a/(1+x) + (b+cx)/(1-x+x^2) = [a(1-x+x^2)+(b+cx)(1+x)]/(1+x^3)
a＋b = 1, -a+b+c = 0, a+c = 0,
a = 1/3， b = 2/3， c = -1/3
I = ∫dx/(1+x^3) = (1/3)∫[1/(1+x) + (2-x)/(1-x+x^2)]dx
= (1/3)ln(1+x) - (1/6)∫[(2x-1-3)/(1-x+x^2)]dx
= (1/3)ln(1+x) - (1/6)∫d(x^2-x+1)/(1-x+x^2) + (1/2)∫dx/[(x-1/2)^2+3/4]
= (1/3)ln(1+x) - (1/6)ln(1-x+x^2) + [(1/2)/(√3/2)]arctan[(x-1/2)/(√3/2)] + C
= (1/3)ln(1+x) - (1/6)ln(1-x+x^2) + (1/√3)arctan[(2x-1)/√3] + C