∫x²+1/(x+1)²(x-1)dx
1个回答
展开全部
(3)
∫(x²+1)/[(x+1)²(x-1)]dx
=∫[x/(x²-1) -1/(x+1)²]dx
=∫[x/(x²-1)]dx -∫[1/(x+1)²]dx
=½∫[2x/(x²-1)]dx - ∫[1/(x+1)²]d(x+1)
=½ln|x²-1| +(x+1)⁻¹ +C
∫(x²+1)/[(x+1)²(x-1)]dx
=∫[x/(x²-1) -1/(x+1)²]dx
=∫[x/(x²-1)]dx -∫[1/(x+1)²]dx
=½∫[2x/(x²-1)]dx - ∫[1/(x+1)²]d(x+1)
=½ln|x²-1| +(x+1)⁻¹ +C
本回答被提问者和网友采纳
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
