高等数学题 18谢谢各位！！
展开全部
∫[xe^x/(1+e^x)^2]dx
换元，令1+e^x=u，则x=ln(u-1)，dx=du/(u-1)
∫[xe^x/(1+e^x)^2]dx
=∫[(u-1)ln(u-1)/(u-1)u^2]du
=∫[ln(u-1)/u^2]du
=-∫ln(u-1)d(1/u)
=-ln(u-1)/u+∫[1/u(u-1)]du
=-ln(u-1)/u+∫[1/(u-1)]du-∫1/udu
=-ln(u-1)/u+ln(u-1)-lnu+C
=-x/(1+e^x)+x-ln(1+e^x)+C
=-x/(1+e^x)+ln[e^x/(1+e^x)]+C
=-x/(1+e^x)-ln[1+e^(-x)]+C
选B
换元，令1+e^x=u，则x=ln(u-1)，dx=du/(u-1)
∫[xe^x/(1+e^x)^2]dx
=∫[(u-1)ln(u-1)/(u-1)u^2]du
=∫[ln(u-1)/u^2]du
=-∫ln(u-1)d(1/u)
=-ln(u-1)/u+∫[1/u(u-1)]du
=-ln(u-1)/u+∫[1/(u-1)]du-∫1/udu
=-ln(u-1)/u+ln(u-1)-lnu+C
=-x/(1+e^x)+x-ln(1+e^x)+C
=-x/(1+e^x)+ln[e^x/(1+e^x)]+C
=-x/(1+e^x)-ln[1+e^(-x)]+C
选B
本回答由提问者推荐
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
