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

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

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

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

Para la integral $$$\int{x \ln{\left(\frac{2 x}{3} \right)} d x}$$$, utiliza la integración por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sean $$$\operatorname{u}=\ln{\left(\frac{2 x}{3} \right)}$$$ y $$$\operatorname{dv}=x dx$$$.

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

Entonces,

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

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

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

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

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

Por lo tanto,

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

Simplificar:

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

Añade la constante de integración:

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

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