∫1/(1+e^x)² dx=∫e^x/[e^x(1+e^x)²] dx=∫1/[e^x(1+e^x)²] de^x令e^x=u=∫1/[u(1+u)²] du=∫(1/u-1/(1+u)-1/(1+u)²] du=ln|u|-ln|u+1|+1/(1+u)+C=x-ln|e^x+1|+1/(e^x+1)+C
