# How do you integrate ln(ln(x))/x?

Aug 7, 2016

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

#### Explanation:

We can first substitute. Let $t = \ln \left(x\right)$ such that $\mathrm{dt} = \frac{1}{x} \mathrm{dx}$. Thus:

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

From here, use integration by parts, which takes the form:

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

So, let:

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

Thus:

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

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

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

$= \ln \left(x\right) \cdot \ln \left(\ln \left(x\right)\right) - \ln \left(x\right)$

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