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

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

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=- \frac{1}{2}$$$ y $$$f{\left(x \right)} = \frac{\ln{\left(- x \right)}}{x^{3}}$$$:

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

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

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

Entonces $$$\operatorname{du}=\left(\ln{\left(- x \right)}\right)^{\prime }dx=\frac{dx}{x}$$$ (los pasos pueden verse ») y $$$\operatorname{v}=\int{\frac{1}{x^{3}} d x}=- \frac{1}{2 x^{2}}$$$ (los pasos pueden verse »).

La integral puede reescribirse como

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

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=- \frac{1}{2}$$$ y $$$f{\left(x \right)} = \frac{1}{x^{3}}$$$:

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

Aplica la regla de la potencia $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=-3$$$:

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

Por lo tanto,

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

Simplificar:

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

Añade la constante de integración:

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

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