Integral de $$$\frac{1}{x^{2} \ln\left(x\right)}$$$

Halla $$$\int \frac{1}{x^{2} \ln\left(x\right)}\, dx$$$.

Sea $$$u=\frac{1}{x}$$$.

Entonces $$$du=\left(\frac{1}{x}\right)^{\prime }dx = - \frac{1}{x^{2}} dx$$$ (los pasos pueden verse »), y obtenemos que $$$\frac{dx}{x^{2}} = - du$$$.

Por lo tanto,

$${\color{red}{\int{\frac{1}{x^{2} \ln{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1}{\ln{\left(u \right)}} d u}}}$$

Esta integral (Integral logarítmica) no tiene una forma cerrada:

$${\color{red}{\int{\frac{1}{\ln{\left(u \right)}} d u}}} = {\color{red}{\operatorname{li}{\left(u \right)}}}$$

Recordemos que $$$u=\frac{1}{x}$$$:

$$\operatorname{li}{\left({\color{red}{u}} \right)} = \operatorname{li}{\left({\color{red}{\frac{1}{x}}} \right)}$$

Por lo tanto,

$$\int{\frac{1}{x^{2} \ln{\left(x \right)}} d x} = \operatorname{li}{\left(\frac{1}{x} \right)}$$

Añade la constante de integración:

$$\int{\frac{1}{x^{2} \ln{\left(x \right)}} d x} = \operatorname{li}{\left(\frac{1}{x} \right)}+C$$

$$$\int \frac{1}{x^{2} \ln\left(x\right)}\, dx = \operatorname{li}{\left(\frac{1}{x} \right)} + C$$$A