Find # int \ ln(lnx)/x \ dx#?

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Answer 1 · 2017-12-04T13:30:34

Substitution Method.

Let #log x=t#
Now differentiate this on both sides.
#implies# #1/x dx=dt#
Now,#int logx# is done by product rule.
Write #logx# as #1*logx#

Answer 2 · 2017-12-04T14:08:23

# int \ ln(lnx)/x \ dx = lnxln(lnx)-lnx + c #

Assuming logarithm base #e#, We seek:

# I = int \ ln(lnx)/x \ dx #

We can perform a substitution:

# u= lnx => (du)/dx=1/x #

Substituting into the integral we get:

# I = int \ lnu \ du #

This is now a standard integral (and can be readily derived with an application of Integration By Parts), and we have:

# I = u lnu-u + c #

And restoring the substitution we have:

# I = lnxln(lnx)-lnx + c #