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:

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

$=tln(t)-intdt$

$=tln(t)-t$

$=ln(x)*ln(ln(x))-ln(x)$

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

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