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

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Answer 1 · 2017-03-07T20:55:42

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

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:

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

$=tlnt-intdt$

$=tlnt-t$

$=t(lnt-1)$

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

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