Integral de $$$\frac{i a g h o r^{3} t w \ln^{2}\left(x\right)}{2 e}$$$ con respecto a $$$x$$$

Halla $$$\int \frac{i a g h o r^{3} t w \ln^{2}\left(x\right)}{2 e}\, dx$$$.

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{i a g h o r^{3} t w}{2 e}$$$ y $$$f{\left(x \right)} = \ln{\left(x \right)}^{2}$$$:

$${\color{red}{\int{\frac{i a g h o r^{3} t w \ln{\left(x \right)}^{2}}{2 e} d x}}} = {\color{red}{\left(\frac{i a g h o r^{3} t w \int{\ln{\left(x \right)}^{2} d x}}{2 e}\right)}}$$

Para la integral $$$\int{\ln{\left(x \right)}^{2} d x}$$$, utiliza la integración por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sean $$$\operatorname{u}=\ln{\left(x \right)}^{2}$$$ y $$$\operatorname{dv}=dx$$$.

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

Entonces,

$$\frac{i a g h o r^{3} t w {\color{red}{\int{\ln{\left(x \right)}^{2} d x}}}}{2 e}=\frac{i a g h o r^{3} t w {\color{red}{\left(\ln{\left(x \right)}^{2} \cdot x-\int{x \cdot \frac{2 \ln{\left(x \right)}}{x} d x}\right)}}}{2 e}=\frac{i a g h o r^{3} t w {\color{red}{\left(x \ln{\left(x \right)}^{2} - \int{2 \ln{\left(x \right)} d x}\right)}}}{2 e}$$

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

$$\frac{i a g h o r^{3} t w \left(x \ln{\left(x \right)}^{2} - {\color{red}{\int{2 \ln{\left(x \right)} d x}}}\right)}{2 e} = \frac{i a g h o r^{3} t w \left(x \ln{\left(x \right)}^{2} - {\color{red}{\left(2 \int{\ln{\left(x \right)} d x}\right)}}\right)}{2 e}$$

Para la integral $$$\int{\ln{\left(x \right)} d x}$$$, 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 $$$\operatorname{dv}=dx$$$.

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

Entonces,

$$\frac{i a g h o r^{3} t w \left(x \ln{\left(x \right)}^{2} - 2 {\color{red}{\int{\ln{\left(x \right)} d x}}}\right)}{2 e}=\frac{i a g h o r^{3} t w \left(x \ln{\left(x \right)}^{2} - 2 {\color{red}{\left(\ln{\left(x \right)} \cdot x-\int{x \cdot \frac{1}{x} d x}\right)}}\right)}{2 e}=\frac{i a g h o r^{3} t w \left(x \ln{\left(x \right)}^{2} - 2 {\color{red}{\left(x \ln{\left(x \right)} - \int{1 d x}\right)}}\right)}{2 e}$$

Aplica la regla de la constante $$$\int c\, dx = c x$$$ con $$$c=1$$$:

$$\frac{i a g h o r^{3} t w \left(x \ln{\left(x \right)}^{2} - 2 x \ln{\left(x \right)} + 2 {\color{red}{\int{1 d x}}}\right)}{2 e} = \frac{i a g h o r^{3} t w \left(x \ln{\left(x \right)}^{2} - 2 x \ln{\left(x \right)} + 2 {\color{red}{x}}\right)}{2 e}$$

Por lo tanto,

$$\int{\frac{i a g h o r^{3} t w \ln{\left(x \right)}^{2}}{2 e} d x} = \frac{i a g h o r^{3} t w \left(x \ln{\left(x \right)}^{2} - 2 x \ln{\left(x \right)} + 2 x\right)}{2 e}$$

Simplificar:

$$\int{\frac{i a g h o r^{3} t w \ln{\left(x \right)}^{2}}{2 e} d x} = \frac{i a g h o r^{3} t w x \left(\ln{\left(x \right)}^{2} - 2 \ln{\left(x \right)} + 2\right)}{2 e}$$

Añade la constante de integración:

$$\int{\frac{i a g h o r^{3} t w \ln{\left(x \right)}^{2}}{2 e} d x} = \frac{i a g h o r^{3} t w x \left(\ln{\left(x \right)}^{2} - 2 \ln{\left(x \right)} + 2\right)}{2 e}+C$$

$$$\int \frac{i a g h o r^{3} t w \ln^{2}\left(x\right)}{2 e}\, dx = \frac{i a g h o r^{3} t w x \left(\ln^{2}\left(x\right) - 2 \ln\left(x\right) + 2\right)}{2 e} + C$$$A