设t=e^x 则dx=dt\tdx\(1+e^2x)^(1\2)=dt\t(1+t^2)^(1\2)又设u=1\t>0 则dt=-du\u^2dx\(1+e^2x)^(1\2)=-du\(u^2+1)^(1\2)∴∫dx\(1+e^2x)^(1\2)=-In[u+(1+u^2)^(1\2)]+C=In{e^x\[1+(1+e^2x)^(1\2)]}+C
