How do you integrate #ln(ln(x))/x#?

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Answer 1 · 2016-08-07T04:46:10

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

We can first substitute. Let #t=ln(x)# such that #dt=1/xdx#. Thus:

#intln(ln(x))/xdx=intln(ln(x))1/xdx=intln(t)dt#

From here, use integration by parts, which takes the form:

#intudv=uv-intvdu#

So, let:

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

Thus:

#intln(t)dt=tln(t)-intt(1/t)dt#

#=tln(t)-intdt#

#=tln(t)-t#

#=ln(x)*ln(ln(x))-ln(x)#

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