求不定积分∫dx/x[根号1-(ln^2)x]
人气:203 ℃ 时间:2019-09-27 15:35:58
解答
∫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
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