What is the indefinite integral of #1/(xlnx)#?

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Answer 1 · 2016-05-30T01:58:40

#ln(abslnx)+C#

We have the integral:

#int1/(xlnx)dx#

Use substitution. Let #u=lnx# so that #du=1/xdx#. Note that both of these are currently present in the integral.

#int1/(xlnx)dx=int(1/lnx)1/xdx=int1/udu#

This is a common integral:

#int1/udu=ln(absu)+C#

Since #u=lnx#:

#ln(absu)+C=ln(abslnx)+C#