求不定积分∫dx⼀(1+x^1⼀2 ) 求高手解题要步骤谢谢

2025-02-25 08:51:12
推荐回答(3个)
回答1:

楼下不负责任呢厄,,

回答2:

令√x=t
则原式=∫2tdt/(1+t)
=∫(2t+2-2)dt/(1+t)
=∫2dt-2∫dt/(1+t)
=2t-2ln(1+t)+C
=2√x-2ln(1+√x)+C

回答3:

令√x=t
x=t^2
dx=2tdt
∫dx/(1+x^1/2 )

=∫1/(1+t)*2tdt
=2∫[1-1/(1+t)]dt
=2t-2ln(1+t)+C
自己反代吧