# What is the integral of [ln(lnx)]/[x] dx?

Apr 26, 2015

I got:

$\ln x \cdot \ln \left(\ln x\right) - \ln x + C$

$f \left(x\right) = \ln \frac{\ln x}{x}$

Now we can do some unusual substitution...

First, recognize that $d \left(\ln x\right) = \frac{1}{x} \mathrm{dx}$. Then, realize that $\frac{1}{x} \mathrm{dx}$ is in there. So, we have:

$f \left(u\right) = \ln \left(u\right) \mathrm{du}$
where $u = \ln x$ and $\mathrm{du} = \frac{1}{x} \mathrm{dx}$.

Further considerations lead to using integration by parts on $\int \ln u \mathrm{du}$. Let:

$s = \ln u$
$\mathrm{ds} = \frac{1}{u} \mathrm{du}$
$\mathrm{dt} = \mathrm{du}$
$t = u$

$\int \ln \left(u\right) \mathrm{du}$

$= s t - \int t \mathrm{ds}$

$= u \cdot \ln u - \int u \cdot \frac{1}{u} \mathrm{du}$

$= u \cdot \ln u - u$

$= \textcolor{b l u e}{\ln x \cdot \ln \left(\ln x\right) - \ln x + C}$