換元整體令√(e^x+1)=t
所以x=ln(t^2-1)
原式=∫tdln(t^2-1)
=∫t*2t/(t^2-1)dt
=∫(2t^2-2+2)/(t^2-1)dt
=∫[2+2/(t^2-1)]dt
=2t|（0,ln2）+∫(1/(t-1)-1/(t+1))dt
=2ln2+ln|(t-1)/(t+1)||(0,ln2)
至于你寫錯了就更簡單了
也是令t=√(e^x-1),t=ln(t^2+1)
原式=∫tdln(t^2+1)
=∫2t^2/(1+t^2)dt
=∫（2t^2+2-2)/(t^2+1)dt
=∫2dt-2∫1/(t^2+1)dt
=2-2arctant|(0,1)
=2-2arctant1
=2-π/2