不定积分求∫[x^3/(1+x^3)]dx
1个回答
展开全部
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)。
=∫[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)。
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
