x的平方+x+1分之x+1的不定积分
2个回答
展开全部
∫(x+1)/(x^2+x+1) dx
=(1/2)∫(2x+1)/(x^2+x+1) dx +(1/2)∫dx/(x^2+x+1)
=(1/2)ln|x^2+x+1| +(1/2)∫dx/[3/4+ (x+1/2)^2]
=(1/2)ln|x^2+x+1| +(2/3)∫dx/{ 1+ [(x+1/2)/(√3/2)]^2}
=(1/2)ln|x^2+x+1| +[4/(3√3) ]∫d[(x+1/2)/(√3/2)]/{ 1+ [(x+1/2)/(√3/2)]^2}
=(1/2)ln|x^2+x+1| +[4/(3√3) ] arctan[(x+1/2)/(√3/2)] +C
=(1/2)∫(2x+1)/(x^2+x+1) dx +(1/2)∫dx/(x^2+x+1)
=(1/2)ln|x^2+x+1| +(1/2)∫dx/[3/4+ (x+1/2)^2]
=(1/2)ln|x^2+x+1| +(2/3)∫dx/{ 1+ [(x+1/2)/(√3/2)]^2}
=(1/2)ln|x^2+x+1| +[4/(3√3) ]∫d[(x+1/2)/(√3/2)]/{ 1+ [(x+1/2)/(√3/2)]^2}
=(1/2)ln|x^2+x+1| +[4/(3√3) ] arctan[(x+1/2)/(√3/2)] +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