题目

求定积分在区间（正无穷~0）∫1/(1+e^x) dx

提问时间：2020-10-31

答案

令y=e^x => x=lny,dx=1/y dy
当x=0,y=1 // 当x->+∞,y->+∞
∫[0,+∞]1/(1+e^x) dx
= ∫[1,+∞]1/[y(1+y)] dy
= ∫[1,+∞][(1+y)-y]/[y(1+y)] dy
= ∫[1,+∞][1/y-1/(1+y)] dy
= ln|y| - ln|1+y|
= ln|y/(1+y)|
= ln|1/(1+1/y)|
= ln[1/(1+0)] - ln[1/(1+1)]
= ln(1) - ln(1/2)
= ln(2)