求不定积分∫根号(e^x+1) dx 注意 根号内是 e^x+1

Answer 1

t=(e^x+1)^0.5
dx=2t/(t^2-1)
∫(e^x+1)^0.5 dx
=∫2t^2/(t^2-1)dt
=∫2 +2/(t^2-1)dt
=2t+ln[(t-1)/(t+1)]+c