①（1+x）y'=2e^(-y) -1则(1+x)dy/dx=2e^(-y) -1dy/[2e^(-y) -1]=dx/(1+x)两边积分得∫dy/[2e^(-y) -1]=∫dx/(1+x)令e^(-y)=t,y=-lnt∫dy/[2e^(-y) -1]=-∫1/t(2t-1)dt=-2∫(1/(2t-1)-1/2t)dt=-∫1/(2t-1)d(2t-1)+∫...