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

Jun 27, 2016

lnx ln ln x- ln x

Explanation:

This can be done by u substitution. Let lnx =u, so that $\frac{1}{x} \mathrm{dx} = \mathrm{du}$

Jun 27, 2016

$\int \frac{\ln \left(\ln x\right)}{x} \mathrm{dx} = \ln x \ln \left(\ln x\right) - \ln x + c$

Explanation:

Let $z = \ln x$ then $\mathrm{dz} = \frac{\mathrm{dx}}{x}$

Hence $\int \frac{\ln \left(\ln x\right)}{x} \mathrm{dx} = \int \ln z \mathrm{dz}$

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

As $\int u \mathrm{dv} = u v - \int v \mathrm{du}$, we have

$\int \ln z \mathrm{dz} = \ln z \times z - \int z \frac{\mathrm{dz}}{z} = z \ln z - z$

Hence $\int \frac{\ln \left(\ln x\right)}{x} \mathrm{dx} = \ln x \ln \left(\ln x\right) - \ln x + c$