求高数大佬 ！！急
1个回答
展开全部
换元u=√(3x+1)，dx=d(u²-1)/3=2u/3du
=∫(1到2)2u/3*1/((u²-1)/3+u)=∫2u/(u²+3u-1)du=∫(2u+3-3)/(u²+3u-1)du
=∫1/(u²+3u-1)d(u²+3u-1)-∫3/(u+3/2+√13/2)(u+3/2-√13/2)du
=ln(u²+3u-1)-3/√13∫(1/(u+3/2-√13/2)-1/(u+3/2+√13/2))du
=ln(u²+3u-1)-3/√13(ln(u+3/2-√13/2)/(u+3/2+√13/2))
=∫(1到2)2u/3*1/((u²-1)/3+u)=∫2u/(u²+3u-1)du=∫(2u+3-3)/(u²+3u-1)du
=∫1/(u²+3u-1)d(u²+3u-1)-∫3/(u+3/2+√13/2)(u+3/2-√13/2)du
=ln(u²+3u-1)-3/√13∫(1/(u+3/2-√13/2)-1/(u+3/2+√13/2))du
=ln(u²+3u-1)-3/√13(ln(u+3/2-√13/2)/(u+3/2+√13/2))
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
