求不定积分∫dx/x[根号1-(ln^2)x]

题目

求不定积分∫dx/x[根号1-(ln^2)x]

答案

∫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