What is #int ln(lnx)/xdx#?

1 Answer
Nov 3, 2015

Consider the following substitution.

Let:
#t = lnx#
#dt = 1/xdx#

#intln(lnx)/(x)dx = intlntdt#

Now we can do an integration by parts.

#int udv = uv - intvdu#

Let:
#u = lnt#
#du = 1/tdt#
#dv = dt#
#v = t#

#=lnt - int t*1/tdt#

#= tlnt - t#

And now let's substitute back in our original variables.

#= color(blue)((lnx)ln(lnx) - lnx + C)#