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

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

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

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

La integral se convierte en

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

Aplica la regla de la potencia $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=- \frac{1}{2}$$$:

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

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

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

Por lo tanto,

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

Añade la constante de integración:

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

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