用换元法求定积分∫01 1⼀1+√x dx

2025-02-25 09:23:12
推荐回答(1个)
回答1:

解:令x=t^2,则dx=2tdt
∫(0,1) 1/(1+√x) dx
=∫(0,1) 2tdt/(1+t)
=∫(0,1) [2-2/(1+t)]dt
=2t|(0,1)-2ln|1+t| |(0,1)
=2-2ln2