∫(0,+∞)xe^x/(1+e^x)^2dx,求出来了,但是感觉不对!用定积分先求出了.最后正无穷怎么带呀!=-∫xd[1/(1+e^x)]=-x/(1+e^x)+∫[1/(1+e^x)]dx=-x/(1+e^x)+∫[(1+e^x-e^x)/(1+e^x)]dx=-x/(1+e^x)+∫1dx-∫(1/(1+e^x))d(1+e^x)=-x/(1+

Question

∫ (0,+∞)xe^x/(1+e^x)^2dx,求出来了,但是感觉不对!
用定积分先求出了.最后正无穷怎么带呀!
= -∫xd[1/(1+e^x)]
= -x/(1+e^x)+∫[1/(1+e^x)]dx
= -x/(1+e^x)+∫[(1+e^x-e^x)/(1+e^x)]dx
= -x/(1+e^x)+∫1dx-∫(1/(1+e^x))d(1+e^x)
=-x/(1+e^x)+x-ln(1+e^x)+c
最后给的参考答案是ln2

Answer 1