Question

∫1/(1+x^3)dx怎么解？

Answer 1

解：原式=∫dx/[(x+1)(x²-x+1)]
=∫[1/(x+1)+1/(x²-x+1)]dx
=∫dx/(x+1)+∫dx/[(x-1/2)²+(√3/2)²]
=ln│x+1│+2√3/3arctan[(2/√3)(x-1/2)]+C (C是积分常数)
=ln│x+1│+2√3/3arctan[(2x-1)/√3]+C。

追问

原式=∫dx/[(x+1)(x²-x+1)]
=∫[1/(x+1)+1/(x²-x+1)]dx
这不等的吧！！！

追答

我把系数写掉了！应该是：
解：原式=∫dx/[(x+1)(x²-x+1)]
=∫[(1/3)/(x+1)+(-x/3+2/3)/(x²-x+1)]dx
=(1/3)ln│x+1│+(1/6)∫(3+1-2x)/(x²-x+1)dx
=(1/3)ln│x+1│+(1/2)∫dx/(x²-x+1)-(1/6)∫(2x-1)/(x²-x+1)dx
=(1/3)ln│x+1│+(1/2)∫dx/[3/4+(x-1/2)²]-(1/6)ln(x²-x+1)
=(1/6)ln[(x+1)²/(x²-x+1)]+(√3/3)∫d((2x-1)/√3)/[1+((2x-1)/√3)²]
=(1/6)ln[(x+1)²/(x²-x+1)]+(√3/3)arctan[(2x-1)/√3]+C (C是积分常数)。