1.令t=(1+e^x ) ^1/2 x=㏑(t^2-1).dx=2t/(t^2-1)dt.
∫dx/(1+e^x)^1/2=∫[2/(t^2-1)]dt=∫[(1+t+1-t)/(t^2-1)]dt=ln|(t-1)/(t+1)|+C,再回代就可以了.
2.令e^x=t--->x=lnt--->dx=dt/t.[e^x+e^(-x)=e^x+1/e^x=t+1/t]
∫dx/[e^x+e^(-x)]
=∫(dt/t)/(t+1/t)
=∫dt/(t^2+1)
=arctant+C
=arctan(e^x)+C
哪里不懂?