What is #int 1/(xlnx)dx#?

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Answer 1 · 2017-02-16T08:21:46

#int dx/(xlnx) = ln abs(lnx)+C#

Evaluate:

#int dx/(xlnx)#

Note that:

#dx/x = d(lnx) #

So:

#int dx/(xlnx) = int (d(lnx))/lnx#

Substituting #t = lnx# we then have:

#int dx/(xlnx) = int (dt)/t = ln abs(t)+C#

And undoing the substitution:

#int dx/(xlnx) = ln abs(lnx)+C#