Integral de $$$\frac{6 \ln^{2}\left(x\right)}{x}$$$

Halla $$$\int \frac{6 \ln^{2}\left(x\right)}{x}\, dx$$$.

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=6$$$ y $$$f{\left(x \right)} = \frac{\ln{\left(x \right)}^{2}}{x}$$$:

$${\color{red}{\int{\frac{6 \ln{\left(x \right)}^{2}}{x} d x}}} = {\color{red}{\left(6 \int{\frac{\ln{\left(x \right)}^{2}}{x} d x}\right)}}$$

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 puede reescribirse como

$$6 {\color{red}{\int{\frac{\ln{\left(x \right)}^{2}}{x} d x}}} = 6 {\color{red}{\int{u^{2} d u}}}$$

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

$$6 {\color{red}{\int{u^{2} d u}}}=6 {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=6 {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

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

$$2 {\color{red}{u}}^{3} = 2 {\color{red}{\ln{\left(x \right)}}}^{3}$$

Por lo tanto,

$$\int{\frac{6 \ln{\left(x \right)}^{2}}{x} d x} = 2 \ln{\left(x \right)}^{3}$$

Añade la constante de integración:

$$\int{\frac{6 \ln{\left(x \right)}^{2}}{x} d x} = 2 \ln{\left(x \right)}^{3}+C$$

$$$\int \frac{6 \ln^{2}\left(x\right)}{x}\, dx = 2 \ln^{3}\left(x\right) + C$$$A