What is the antiderivative of #ln(x)/x#?

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Answer 1 · 2017-02-21T04:40:19

#ln^2(x)/2 +C#

#I=int lnx/x dx#

Let #u=lnx#

#(du)/dx = 1/x -> dx=x du# [Standard differential]

#:. I= int u/x * x du = int u du#

#= u^2/2 # [Power rule]

Undo u substitution:

#I = ln^2 (x)/2 +C#