∫1/（1+X的三次方）dx怎么做啊啊啊啊啊 我用立方公式了啊…可是怎么才能抵消啊啊啊啊

∫ 1/(1+x³) dx
=∫ 1/[(1+x)(x²-x+1)] dx
=(1/3)∫ 1/(x+1) dx - (1/3)∫ (x-2)/(x²-x+1) dx
=(1/3)ln|x+1| - (1/6)∫ (2x-1-3)/(x²-x+1) dx
=(1/3)ln|x+1| - (1/6)∫ (2x-1)/(x²-x+1) dx + (1/2)∫ 1/(x²-x+1) dx
=(1/3)ln|x+1| - (1/6)∫ 1/(x²-x+1) d(x²-x) + (1/2)∫ 1/[(x-1/2)²+3/4] dx
=(1/3)ln|x+1| - (1/6)ln(x²-x+1) + (√3/3)arctan((2x-1)/√3) + C
已用数学软件验算过,结果正确.
希望可以帮到你,不明白可以追问,如果解决了问题,请点下面的"选为满意回答"按钮,谢谢.