Integral de $$$x^{2} \left(6 - x^{3}\right)^{5}$$$

Halla $$$\int x^{2} \left(6 - x^{3}\right)^{5}\, dx$$$.

Sea $$$u=6 - x^{3}$$$.

Entonces $$$du=\left(6 - x^{3}\right)^{\prime }dx = - 3 x^{2} dx$$$ (los pasos pueden verse »), y obtenemos que $$$x^{2} dx = - \frac{du}{3}$$$.

La integral se convierte en

$${\color{red}{\int{x^{2} \left(6 - x^{3}\right)^{5} d x}}} = {\color{red}{\int{\left(- \frac{u^{5}}{3}\right)d u}}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=- \frac{1}{3}$$$ y $$$f{\left(u \right)} = u^{5}$$$:

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

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

$$- \frac{{\color{red}{\int{u^{5} d u}}}}{3}=- \frac{{\color{red}{\frac{u^{1 + 5}}{1 + 5}}}}{3}=- \frac{{\color{red}{\left(\frac{u^{6}}{6}\right)}}}{3}$$

Recordemos que $$$u=6 - x^{3}$$$:

$$- \frac{{\color{red}{u}}^{6}}{18} = - \frac{{\color{red}{\left(6 - x^{3}\right)}}^{6}}{18}$$

Por lo tanto,

$$\int{x^{2} \left(6 - x^{3}\right)^{5} d x} = - \frac{\left(6 - x^{3}\right)^{6}}{18}$$

Simplificar:

$$\int{x^{2} \left(6 - x^{3}\right)^{5} d x} = - \frac{\left(x^{3} - 6\right)^{6}}{18}$$

Añade la constante de integración:

$$\int{x^{2} \left(6 - x^{3}\right)^{5} d x} = - \frac{\left(x^{3} - 6\right)^{6}}{18}+C$$

$$$\int x^{2} \left(6 - x^{3}\right)^{5}\, dx = - \frac{\left(x^{3} - 6\right)^{6}}{18} + C$$$A