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

2 Answers
Dec 4, 2017

Substitution Method.

Explanation:

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#

Dec 4, 2017

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

Explanation:

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 #