Integral de $$$- \frac{2 \ln\left(3 x\right)}{x^{5}}$$$

Halla $$$\int \left(- \frac{2 \ln\left(3 x\right)}{x^{5}}\right)\, dx$$$.

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=-2$$$ y $$$f{\left(x \right)} = \frac{\ln{\left(3 x \right)}}{x^{5}}$$$:

$${\color{red}{\int{\left(- \frac{2 \ln{\left(3 x \right)}}{x^{5}}\right)d x}}} = {\color{red}{\left(- 2 \int{\frac{\ln{\left(3 x \right)}}{x^{5}} d x}\right)}}$$

Para la integral $$$\int{\frac{\ln{\left(3 x \right)}}{x^{5}} d x}$$$, utiliza la integración por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sean $$$\operatorname{u}=\ln{\left(3 x \right)}$$$ y $$$\operatorname{dv}=\frac{dx}{x^{5}}$$$.

Entonces $$$\operatorname{du}=\left(\ln{\left(3 x \right)}\right)^{\prime }dx=\frac{dx}{x}$$$ (los pasos pueden verse ») y $$$\operatorname{v}=\int{\frac{1}{x^{5}} d x}=- \frac{1}{4 x^{4}}$$$ (los pasos pueden verse »).

La integral se convierte en

$$- 2 {\color{red}{\int{\frac{\ln{\left(3 x \right)}}{x^{5}} d x}}}=- 2 {\color{red}{\left(\ln{\left(3 x \right)} \cdot \left(- \frac{1}{4 x^{4}}\right)-\int{\left(- \frac{1}{4 x^{4}}\right) \cdot \frac{1}{x} d x}\right)}}=- 2 {\color{red}{\left(- \int{\left(- \frac{1}{4 x^{5}}\right)d x} - \frac{\ln{\left(3 x \right)}}{4 x^{4}}\right)}}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=- \frac{1}{4}$$$ y $$$f{\left(x \right)} = \frac{1}{x^{5}}$$$:

$$2 {\color{red}{\int{\left(- \frac{1}{4 x^{5}}\right)d x}}} + \frac{\ln{\left(3 x \right)}}{2 x^{4}} = 2 {\color{red}{\left(- \frac{\int{\frac{1}{x^{5}} d x}}{4}\right)}} + \frac{\ln{\left(3 x \right)}}{2 x^{4}}$$

Aplica la regla de la potencia $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=-5$$$:

$$- \frac{{\color{red}{\int{\frac{1}{x^{5}} d x}}}}{2} + \frac{\ln{\left(3 x \right)}}{2 x^{4}}=- \frac{{\color{red}{\int{x^{-5} d x}}}}{2} + \frac{\ln{\left(3 x \right)}}{2 x^{4}}=- \frac{{\color{red}{\frac{x^{-5 + 1}}{-5 + 1}}}}{2} + \frac{\ln{\left(3 x \right)}}{2 x^{4}}=- \frac{{\color{red}{\left(- \frac{x^{-4}}{4}\right)}}}{2} + \frac{\ln{\left(3 x \right)}}{2 x^{4}}=- \frac{{\color{red}{\left(- \frac{1}{4 x^{4}}\right)}}}{2} + \frac{\ln{\left(3 x \right)}}{2 x^{4}}$$

Por lo tanto,

$$\int{\left(- \frac{2 \ln{\left(3 x \right)}}{x^{5}}\right)d x} = \frac{\ln{\left(3 x \right)}}{2 x^{4}} + \frac{1}{8 x^{4}}$$

Simplificar:

$$\int{\left(- \frac{2 \ln{\left(3 x \right)}}{x^{5}}\right)d x} = \frac{4 \ln{\left(x \right)} + 1 + 4 \ln{\left(3 \right)}}{8 x^{4}}$$

Añade la constante de integración:

$$\int{\left(- \frac{2 \ln{\left(3 x \right)}}{x^{5}}\right)d x} = \frac{4 \ln{\left(x \right)} + 1 + 4 \ln{\left(3 \right)}}{8 x^{4}}+C$$

$$$\int \left(- \frac{2 \ln\left(3 x\right)}{x^{5}}\right)\, dx = \frac{4 \ln\left(x\right) + 1 + 4 \ln\left(3\right)}{8 x^{4}} + C$$$A