大学高数的定积分的问题 50
2个回答
展开全部
∫1/(x²-a²)²dx=1/2∫1/[x(x²-a²)²]d(x²-a²)=-1/2∫1/xd1/(x²-a²)
=-1/[2x(x²-a²)]+1/2∫1/(x²-a²)d1/x
=-1/[2x(x²-a²)]-1/2∫1/[x²(x²-a²)]dx
=-1/[2x(x²-a²)]-1/(2a²)∫1/(x²-a²)-1/x²dx
=-1/[2x(x²-a²)]-1/(4a³)∫1/(x-a)-1/(x+a)dx+1/(2a²)∫1/x²dx
=-1/[2x(x²-a²)]-1/(4a³)ln|(x-a)/(x+a)|-1/(2a²x)+C
=-1/[2x(x²-a²)]+1/2∫1/(x²-a²)d1/x
=-1/[2x(x²-a²)]-1/2∫1/[x²(x²-a²)]dx
=-1/[2x(x²-a²)]-1/(2a²)∫1/(x²-a²)-1/x²dx
=-1/[2x(x²-a²)]-1/(4a³)∫1/(x-a)-1/(x+a)dx+1/(2a²)∫1/x²dx
=-1/[2x(x²-a²)]-1/(4a³)ln|(x-a)/(x+a)|-1/(2a²x)+C
本回答由网友推荐
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
