∫x/(x²+x+1)dx
展开全部
∫[x/(x²+x+1)]dx = (1/2)∫[(2x+1-1)/(x²+x+1)]dx
= (1/2)∫[(2x+1)/(x²+x+1)]dx - (1/2)∫d(x+1/2)/[(x+1/2)²+3/4]
= (1/2)ln(x²+x+1) - (1/2)(2/√3)arctan[(2x+1)/√3] + C
= (1/2)ln(x²+x+1) - (1/√3)arctan[(2x+1)/√3] + C
= (1/2)∫[(2x+1)/(x²+x+1)]dx - (1/2)∫d(x+1/2)/[(x+1/2)²+3/4]
= (1/2)ln(x²+x+1) - (1/2)(2/√3)arctan[(2x+1)/√3] + C
= (1/2)ln(x²+x+1) - (1/√3)arctan[(2x+1)/√3] + C
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
