How do you integrate $[ln(lnx)]/[x] dx$ ?

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Answer 1 · 2016-06-27T04:54:59

lnx ln ln x- ln x

This can be done by u substitution. Let lnx =u, so that $1/x dx =du$

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Answer 2 · 2016-06-27T04:58:44

$int(ln(lnx))/xdx=lnxln(lnx)-lnx+c$

Let $z=lnx$ then $dz=dx/x$

Hence $int(ln(lnx))/xdx=intlnzdz$

Now using integration by parts, if $u=lnz$ and $v=z$

As $intudv=uv-intvdu$ , we have

$intlnzdz=lnzxxz-intzdz/z=zlnz-z$

Hence $int(ln(lnx))/xdx=lnxln(lnx)-lnx+c$

How do you integrate [ln(lnx)]/[x] dx [ln(lnx)]/[x] dx?