这个要自己用待定系数去配。因为1+x^3=(1+x)(1-x+x^2)
所以
先令1/(1+x^3)=A/(1+x)+(Bx+C)/(1-x+x^2)
通过
通分化简对比左右两边分子得:A+B=0,-A+B+C=0,A+C=1
求得A=1/3,B=-1/3,C=2/3
所以,∫[1/(1+x^3)]dx=(1/3)∫[1/(1+x)]dx+∫[(-x/3+2/3)/(1-x+x^2)]dx
=(1/3)∫[1/(1+x)]dx-(1/6)∫[(2x-1)/(1-x+x^2)]dx+(1/2)∫[1/(1-x+x^2)]dx
=(1/3)ln|x+1|-(1/6)ln|x^2-x+1|+(1/√3)arctan[(2x-1)/√3]+C
=(1/3)ln[|x+1|/√(x^2-x+1)]+(1/√3)arctan[(2x-1)/√3]+C