How do you integrate $\frac{ln (ln x)}{x} d x$ $[ln(lnx)]/[x] dx$ ?

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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 $\frac{1}{x} d x = d u$ $1/x dx =du$

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Answer 2 · 2016-06-27T04:58:44

$\int \frac{ln (ln x)}{x} d x = ln x ln (ln x) - ln x + c$ $int(ln(lnx))/xdx=lnxln(lnx)-lnx+c$

Let $z = ln x$ $z=lnx$ then $d z = \frac{d x}{x}$ $dz=dx/x$

Hence $\int \frac{ln (ln x)}{x} d x = \int ln z d z$ $int(ln(lnx))/xdx=intlnzdz$

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

As $\int u d v = u v - \int v d u$ $intudv=uv-intvdu$ , we have

$\int ln z d z = ln z \times z - \int z \frac{d z}{z} = z ln z - z$ $intlnzdz=lnzxxz-intzdz/z=zlnz-z$

Hence $\int \frac{ln (ln x)}{x} d x = ln x ln (ln x) - ln x + c$ $int(ln(lnx))/xdx=lnxln(lnx)-lnx+c$

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