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

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

Para la integral $$$\int{x^{3} \ln{\left(x \right)} d x}$$$, 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 $$$\operatorname{dv}=x^{3} dx$$$.

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

Entonces,

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

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{1}{4}$$$ y $$$f{\left(x \right)} = x^{3}$$$:

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

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

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

Por lo tanto,

$$\int{x^{3} \ln{\left(x \right)} d x} = \frac{x^{4} \ln{\left(x \right)}}{4} - \frac{x^{4}}{16}$$

Simplificar:

$$\int{x^{3} \ln{\left(x \right)} d x} = \frac{x^{4} \left(4 \ln{\left(x \right)} - 1\right)}{16}$$

Añade la constante de integración:

$$\int{x^{3} \ln{\left(x \right)} d x} = \frac{x^{4} \left(4 \ln{\left(x \right)} - 1\right)}{16}+C$$

$$$\int x^{3} \ln\left(x\right)\, dx = \frac{x^{4} \left(4 \ln\left(x\right) - 1\right)}{16} + C$$$A