1个回答
展开全部
1/[ (u-1)/(1+e^(-u)) - u ]
= 1/[ e^u.(u-1)/(1+e^u) - u ]
=(1+e^u)/[ e^u.(u-1) -u -u.e^u]
=(1+e^u)/( -e^u -u )
=-(1+e^u)/( u+e^u)
∫ du/[ (u-1)/(1+e^(-u)) - u ] = ∫ dy/y
∫ (1+e^u)/( u+e^u) du = -∫ dy/y
ln|u+e^u| = -ln|y| + C'
u+e^u = C/y
= 1/[ e^u.(u-1)/(1+e^u) - u ]
=(1+e^u)/[ e^u.(u-1) -u -u.e^u]
=(1+e^u)/( -e^u -u )
=-(1+e^u)/( u+e^u)
∫ du/[ (u-1)/(1+e^(-u)) - u ] = ∫ dy/y
∫ (1+e^u)/( u+e^u) du = -∫ dy/y
ln|u+e^u| = -ln|y| + C'
u+e^u = C/y
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