How do you evaluate the integral #int (ln(lnx))/x dx#?

1 Answer
mason m
Mar 7, 2017

#lnx(ln(lnx)-1)+C#

Explanation:

First let #t=lnx#. This implies that #dt=1/xdx#. Then:

#intln(lnx)/xdx=intln(lnx)1/xdx=intln(t)dt#

Now we can use integration by parts which takes the form #intudv=uv-intvdu#. Let:

#{(u=lnt" "=>" "du=1/tdt),(dv=dt" "=>" "v=t):}#

Then:

#=intln(t)dt=tlnt-intt1/tdt#

#=tlnt-intdt#

#=tlnt-t#

#=t(lnt-1)#

#=lnx(ln(lnx)-1)+C#

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