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

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

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

Então $$$du=\left(\ln{\left(x \right)}\right)^{\prime }dx = \frac{dx}{x}$$$ (veja os passos »), e obtemos $$$\frac{dx}{x} = du$$$.

Logo,

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

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

Então $$$dv=\left(\ln{\left(u \right)}\right)^{\prime }du = \frac{du}{u}$$$ (veja os passos »), e obtemos $$$\frac{du}{u} = dv$$$.

Logo,

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

Aplique a regra da potência $$$\int v^{n}\, dv = \frac{v^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=1$$$:

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

Recorde que $$$v=\ln{\left(u \right)}$$$:

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

Recorde que $$$u=\ln{\left(x \right)}$$$:

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

Portanto,

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

Adicione a constante de integração:

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

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