求不定积分∫dx/x[根号1-(ln^2)x]
问题描述：
求不定积分∫dx/x[根号1-(ln^2)x]
答
∫dx/x[根号1-(ln^2)x]
=∫dlnx/[根号1-(ln^2)x]
=ln|(1+根号(1-ln^2x))/lnx|+C
答
不给过程
答
∫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
答
∫dx/x√(1-ln²x)
=∫dlnx/√(1-ln²x)
=sin(lnx)+C