Question

求积分∫x/1+x³dx？

修改一下，是∫x/(1+x³)dx

Answer 1

简单计算一下，答案如图所示

Answer 2

解：原式=lnx＋¼x的4次方＋C
直接求解即可

追问

可能我写的不清楚，我修改了一下，你再看一下哈

Answer 3

这是详细过程

Answer 4

设 x/(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 = 0, -A+B+C = 1, A+C = 0
联立解得 A = -1/3， B = C = 1/3
∫[x/(1+x^3)]dx = (1/3)∫[(1+x)/(1-x+x^2)-1/(1+x)]dx
= (1/3)∫(1+x)dx/(1-x+x^2) - (1/3)∫dx/(1+x)
= (1/3)∫(x-1/2+3/2)dx/(x^2-x+1) - (1/3)∫d(1+x)/(1+x)
= (1/6)∫(2x-1)dx/(x^2-x+1) + (1/2)∫d(x-1/2)/[(x-1/2)^2+3/4] - (1/3)ln(1+x)
= (1/6)ln(x^2-x+1) + (1/2)(2/√3)arctan(x-1/2)/(√3/2)] - (1/3)ln(1+x) + C
= (1/6)ln[(x^2-x+1)/(1+x)^2] + (1/√3)arctan[(2x-1)/√3] + C