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

Mar 7, 2017

$\ln x \left(\ln \left(\ln x\right) - 1\right) + C$

#### Explanation:

First let $t = \ln x$. This implies that $\mathrm{dt} = \frac{1}{x} \mathrm{dx}$. Then:

$\int \ln \frac{\ln x}{x} \mathrm{dx} = \int \ln \left(\ln x\right) \frac{1}{x} \mathrm{dx} = \int \ln \left(t\right) \mathrm{dt}$

Now we can use integration by parts which takes the form $\int u \mathrm{dv} = u v - \int v \mathrm{du}$. Let:

$\left\{\begin{matrix}u = \ln t \text{ "=>" "du=1/tdt \\ dv=dt" "=>" } v = t\end{matrix}\right.$

Then:

$= \int \ln \left(t\right) \mathrm{dt} = t \ln t - \int t \frac{1}{t} \mathrm{dt}$

$= t \ln t - \int \mathrm{dt}$

$= t \ln t - t$

$= t \left(\ln t - 1\right)$

$= \ln x \left(\ln \left(\ln x\right) - 1\right) + C$