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

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

Sea $$$u=2 x^{2} - 3$$$.

Entonces $$$du=\left(2 x^{2} - 3\right)^{\prime }dx = 4 x dx$$$ (los pasos pueden verse »), y obtenemos que $$$x dx = \frac{du}{4}$$$.

Por lo tanto,

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

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

$${\color{red}{\int{\frac{u e^{3}}{4} d u}}} = {\color{red}{\left(\frac{e^{3} \int{u d u}}{4}\right)}}$$

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

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

Recordemos que $$$u=2 x^{2} - 3$$$:

$$\frac{e^{3} {\color{red}{u}}^{2}}{8} = \frac{e^{3} {\color{red}{\left(2 x^{2} - 3\right)}}^{2}}{8}$$

Por lo tanto,

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

Añade la constante de integración:

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

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