設(shè)根號(hào)(e^x-1) =t
t^2 +1=e^x
x=ln(t^2 +1)
代入得
∫t dln(t^2 +1)
=∫2t^2/(t^2 +1) dt
=2*∫t^2/(t^2 +1) dt
=2*∫(t^2 +1-1)/(t^2 +1) dt
=2*∫[1 -1/(t^2 +1)] dt
=2*[∫1 dt -∫1/(t^2 +1) dt
=2*(t -arctant) +C(常數(shù))
=2*【(e^x-1) -arctan(e^x-1)】+C
=2*【e^x -arctan(e^x-1)】+C(常數(shù)都?xì)w納到C)