求积分∫ 1⼀(1+e^2x) dx

2024-11-30 05:37:43
推荐回答(1个)
回答1:

设t=e^(2x),x=(lnt)/2,dx=1/(2t) dt
∫dx/[1+e^(2x)]
= (1/2)∫dt/[t(1+t)]
= (1/2)∫[(1+t)-t]/[t(1+t)] dt
= (1/2)∫[1/t - 1/(1+t)] dt
= (1/2)[ln|t| - ln|1+t|] + C
= (1/2)[ln|e^(2x)| - ln|1+e^(2x)] + C
= x - (1/2)ln|1+e^(2x)| + C