∫dx/{[(x+1)^2][(x-1)^4]}^(1/3)的不定积分积分
展开全部
∫dx/[(x+1)²(x-1)⁴]^(1/3)的不定积分积分
原式=∫dx/{[(x+1)^(2/3)][(x-1)^(4/3)]}
=∫dx/{[(x+1)^(2/3)][(x-1)^(-2/3)](x-1)²}
=∫{[(x-1)/(x+1)]^(2/3)}dx/(x-1)²
=-(1/2)∫{(x+1)/(x-1)]^(-2/3)}d[(x+1)/(x-1)]
=-(1/2)(3)[(x+1)/(x-1)]^(1/3)+C
=-(3/2)[(x+1)/(x-1)]^(1/3)+C
原式=∫dx/{[(x+1)^(2/3)][(x-1)^(4/3)]}
=∫dx/{[(x+1)^(2/3)][(x-1)^(-2/3)](x-1)²}
=∫{[(x-1)/(x+1)]^(2/3)}dx/(x-1)²
=-(1/2)∫{(x+1)/(x-1)]^(-2/3)}d[(x+1)/(x-1)]
=-(1/2)(3)[(x+1)/(x-1)]^(1/3)+C
=-(3/2)[(x+1)/(x-1)]^(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
