1个回答
展开全部
令 √(2x+1) = t, 则 x = (t^2-1)/2, dx = tdt
I = (1/4)∫(t^2-1)^2 · t · tdt = (1/4)∫(t^6 - 2t^4 + t^2)dt
= (1/4)[t^7/7 - (2/5)t^5 + t^3/3] + C
= (1/420)t^3(15t^4 - 42t^2 + 35) + C
= (1/105)(15x^2-6x+2)(2x+1)^(3/2) + C
I = (1/4)∫(t^2-1)^2 · t · tdt = (1/4)∫(t^6 - 2t^4 + t^2)dt
= (1/4)[t^7/7 - (2/5)t^5 + t^3/3] + C
= (1/420)t^3(15t^4 - 42t^2 + 35) + C
= (1/105)(15x^2-6x+2)(2x+1)^(3/2) + C
本回答被网友采纳
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
