求不定积分∫(dx)/√(1+e^2x)=?

答：
设t=√ [1+e^(2x) ]
t^2=1+e^(2x)
e^(2x)=t^2-1
2x=ln(t^2-1)
x=(1/2)ln(t^2-1)
原式
=∫ 1/t d [(1/2)ln(t^2-1)]
=(1/2) ∫ (1/t)*[ 1/(t^2-1) ]*2t dt
=∫ 1/(t^2-1) dt
=∫ 1/[(t-1)(t+1)] dt
=(1/2)* ∫ 1/(t-1) -1/(t+1) dt
=(1/2)* [ln(t-1) -ln(t+1) ] +C
=(1/2) ln [(t-1)/(t+1)]+C
=(1/2)*ln { [√(1+e^2x) -1 ] / [ √(1+e^2x) +1 ] } +C