This is an impossible integral to complete without resorting to calculators and evaluating at explicit bounds.
http://www.wolframalpha.com/input/?i=integral+of+1%2F%28ln%28lnx%29%29
Let's see how far we can go, though...
#int (ln(lnx))^(-1)dx#
Let:
#u = lnx#
#du = 1/xdx#
#xdu = dx -> e^udu = dx#
#= int (lnu)^(-1)e^udu#
#= inte^u/(lnu)du#
#st - int t ds#
Let:
#s = e^u#
#ds = e^udu#
#dt = 1/lnudu#
#t = ?#
Detour. We need to know the integral for #1/lnx#. Let:
#q = 1/lnu#
#dq = -lnu/udu#
#dr = du#
#r = u#
#qr - int r dq#
#= u/lnu + int lnudu#
#= u/lnu + u ln|u| - u = t#
Back to #s# and #t#:
#= e^u (u/lnu + u ln|u| - u) - int e^u(u/lnu + u lnu - u)du#
#= (ue^u)/lnu + ue^u ln|u| - u e^u - int (ue^u)/lnudu - int ue^u lnudu + intu e^udu#
The only integral we can do with real functions or standard functions is #int u e^udu#. I'm running out of variables.
#op - int p do#
Let:
#o = u#
#do = du#
#dp = e^udu#
#p = e^u#
#ue^u - int e^udu#
#= ue^u - e^u#
So now we get:
#= (ue^u)/lnu + ue^u ln|u| cancel(- u e^u + ue^u) - e^u - int (ue^u)/lnudu - int ue^u lnudu#
#= (ue^u)/lnu + ue^u ln|u| - e^u - int (ue^u)/lnudu - int ue^u lnudu#
#= e^u[u/lnu + u ln|u| - 1] - int (ue^u)/lnudu - int ue^u lnudu#
#= e^(lnx)[(lnx)/ln(lnx) + (lnx) ln|lnx| - 1] - int ((lnx)e^(lnx))/(xln(lnx))dx - int 1/x(lnx)e^(lnx) ln(lnx)dx#
#= color(blue)(x[(lnx)/ln(lnx) + (lnx) ln|lnx| - 1] - int (lnx)/(ln(lnx))dx - int (lnx) ln(lnx)dx)#
