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

1 Answer
Feb 16, 2017

Answer:

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

Explanation:

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#