求不定积分∫dx⼀x[根号1-(ln^2)x] 需要过程~

2024-11-24 07:29:29
推荐回答(4个)
回答1:

∫dx/x[根号1-(ln^2)x]
=∫d(lnx)/[根号1-(ln^2)x]
=∫dt/[根号1-t^2] (设t=lnx)
=arcsint+C
=arcsin(lnx)+C

回答2:

∫dx/x[根号1-(ln^2)x]
=∫dlnx/[根号1-(ln^2)x]
=ln|(1+根号(1-ln^2x))/lnx|+C

回答3:

∫dx/x√(1-ln²x)
=∫dlnx/√(1-ln²x)
=sin(lnx)+C

回答4:

不给过程