Integral of $$$\frac{\ln\left(\ln\left(x\right)\right)}{x \ln\left(x\right)}$$$

Find $$$\int \frac{\ln\left(\ln\left(x\right)\right)}{x \ln\left(x\right)}\, dx$$$.

Let $$$u=\ln{\left(x \right)}$$$.

Then $$$du=\left(\ln{\left(x \right)}\right)^{\prime }dx = \frac{dx}{x}$$$ (steps can be seen »), and we have that $$$\frac{dx}{x} = du$$$.

The integral can be rewritten as

$${\color{red}{\int{\frac{\ln{\left(\ln{\left(x \right)} \right)}}{x \ln{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\ln{\left(u \right)}}{u} d u}}}$$

Let $$$v=\ln{\left(u \right)}$$$.

Then $$$dv=\left(\ln{\left(u \right)}\right)^{\prime }du = \frac{du}{u}$$$ (steps can be seen »), and we have that $$$\frac{du}{u} = dv$$$.

So,

$${\color{red}{\int{\frac{\ln{\left(u \right)}}{u} d u}}} = {\color{red}{\int{v d v}}}$$

Apply the power rule $$$\int v^{n}\, dv = \frac{v^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:

$${\color{red}{\int{v d v}}}={\color{red}{\frac{v^{1 + 1}}{1 + 1}}}={\color{red}{\left(\frac{v^{2}}{2}\right)}}$$

Recall that $$$v=\ln{\left(u \right)}$$$:

$$\frac{{\color{red}{v}}^{2}}{2} = \frac{{\color{red}{\ln{\left(u \right)}}}^{2}}{2}$$

Recall that $$$u=\ln{\left(x \right)}$$$:

$$\frac{\ln{\left({\color{red}{u}} \right)}^{2}}{2} = \frac{\ln{\left({\color{red}{\ln{\left(x \right)}}} \right)}^{2}}{2}$$

Therefore,

$$\int{\frac{\ln{\left(\ln{\left(x \right)} \right)}}{x \ln{\left(x \right)}} d x} = \frac{\ln{\left(\ln{\left(x \right)} \right)}^{2}}{2}$$

Add the constant of integration:

$$\int{\frac{\ln{\left(\ln{\left(x \right)} \right)}}{x \ln{\left(x \right)}} d x} = \frac{\ln{\left(\ln{\left(x \right)} \right)}^{2}}{2}+C$$

$$$\int \frac{\ln\left(\ln\left(x\right)\right)}{x \ln\left(x\right)}\, dx = \frac{\ln^{2}\left(\ln\left(x\right)\right)}{2} + C$$$A