Question

求定积分∫√(e^x-1)dx (上限㏑2,下限0)求具体过程,

Answer 1

简单计算一下即可，答案如图所示

Answer 2

令t=√(e^x-1),则t^2+1=e^x,换元变积分限,∫tdln(t^2+1) = 2∫t^2/(t^2+1)dt= 2∫dt+ 2∫1/(t^2+1)dt