∫1⼀(1+√x) dx 用第二类换元法求不定积分过程,麻烦高手,谢谢啦

2025-03-07 08:13:41
推荐回答(2个)
回答1:

令√x=t,则x=t^2
dx=d(t^2)=2tdt
∴原式=∫1/(1+√x) dx
=∫2t/(1+t) dt
=∫(2(t+1)-2)/(1+t)dt
=∫2dt-∫2/(1+t)dt
=2t-2ln|t+1|+C
=2√x-2ln|√x+1|+C

回答2:

2√x-2ln|√x+1|+C