∫1/（1+X³）dx 求积分
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原式=(1/3)∫[1/(x+1)+(2-x)/(x²-x+1)]dx
=(1/3)∫[1/(x+1)+(1/2)(3-(2x-1))/(x²-x+1)]dx
=(1/3)∫dx/(x+1)+(1/2)∫dx/(x²-x+1)-(1/6)∫(2x-1)dx/(x²-x+1)
=(1/3)∫dx/(x+1)+(1/√3)∫d((2x-1)/√3)/(1+((2x-1)/√3)²)-(1/6)∫d(x²-x+1)/(x²-x+1)
=(1/3)ln│x+1│+(1/√3)arctan((2x-1)/√3)-(1/6)ln(x²-x+1)+C (C是积分常数)
=(1/3)∫[1/(x+1)+(1/2)(3-(2x-1))/(x²-x+1)]dx
=(1/3)∫dx/(x+1)+(1/2)∫dx/(x²-x+1)-(1/6)∫(2x-1)dx/(x²-x+1)
=(1/3)∫dx/(x+1)+(1/√3)∫d((2x-1)/√3)/(1+((2x-1)/√3)²)-(1/6)∫d(x²-x+1)/(x²-x+1)
=(1/3)ln│x+1│+(1/√3)arctan((2x-1)/√3)-(1/6)ln(x²-x+1)+C (C是积分常数)
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