不定积分求∫[x^3/(1+x^3)]dx

发布网友发布时间：2023-07-27 23:41

共1个回答

热心网友时间：2024-11-11 19:27

解：原式=∫[1-1/((x+1)(x²-x+1))]dx
=∫[1-(1/3)/(x+1)+(x/3-2/3)/(x²-x+1)]dx
=x-ln│x+1│/3+(1/6)∫(2x-1-3)/(x²-x+1)dx
=x-ln│x+1│/3+(1/6)∫(2x-1)/(x²-x+1)dx-(1/2)∫dx/((x-1/2)²+3/4)
=x-ln│x+1│/3+(1/6)∫d(x²-x+1)/(x²-x+1)-(√3/3)∫d[2(x-1/2)/√3]/[1+(2(x-1/2)/√3)²]
=x-ln│x+1│/3+(1/6)ln(x²-x+1)-(√3/3)arctan[2(x-1/2)/√3]+C (C是积分常数)。

热心网友时间：2024-11-11 19:27

解：原式=∫[1-1/((x+1)(x²-x+1))]dx
=∫[1-(1/3)/(x+1)+(x/3-2/3)/(x²-x+1)]dx
=x-ln│x+1│/3+(1/6)∫(2x-1-3)/(x²-x+1)dx
=x-ln│x+1│/3+(1/6)∫(2x-1)/(x²-x+1)dx-(1/2)∫dx/((x-1/2)²+3/4)
=x-ln│x+1│/3+(1/6)∫d(x²-x+1)/(x²-x+1)-(√3/3)∫d[2(x-1/2)/√3]/[1+(2(x-1/2)/√3)²]
=x-ln│x+1│/3+(1/6)ln(x²-x+1)-(√3/3)arctan[2(x-1/2)/√3]+C (C是积分常数)。

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