1个回答
展开全部
∫3/(x^3+1)dx
=∫(x^2-x+1)+(x+1)-(x^2-1)/(x+1)(x^2-x+1)
=∫dx/(x+1)+∫dx/(x^2-x+1)-∫(x-1)/(x^2-x+1)dx
=∫dx/(x+1)-1/2∫(2x-1-1)/(x^2-x+1)+∫dx/(x^2-x+1)
=∫dx/(x+1)-1/2∫(x^2-x+1)/(x^2-x+1)+3/2∫dx/(x^2-x+1)
=ln|x+1|-(1/2)ln|x^2-x+1|+√3 arctan[(2x-1)/√3]+c
=∫(x^2-x+1)+(x+1)-(x^2-1)/(x+1)(x^2-x+1)
=∫dx/(x+1)+∫dx/(x^2-x+1)-∫(x-1)/(x^2-x+1)dx
=∫dx/(x+1)-1/2∫(2x-1-1)/(x^2-x+1)+∫dx/(x^2-x+1)
=∫dx/(x+1)-1/2∫(x^2-x+1)/(x^2-x+1)+3/2∫dx/(x^2-x+1)
=ln|x+1|-(1/2)ln|x^2-x+1|+√3 arctan[(2x-1)/√3]+c
本回答由提问者推荐
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
