Integral de $$$\frac{e^{\frac{1}{x^{3}}}}{x^{4}}$$$

Halla $$$\int \frac{e^{\frac{1}{x^{3}}}}{x^{4}}\, dx$$$.

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

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

Entonces,

$${\color{red}{\int{\frac{e^{\frac{1}{x^{3}}}}{x^{4}} d x}}} = {\color{red}{\int{\frac{e^{\frac{1}{u}}}{3 u^{2}} d u}}}$$

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 $$$f{\left(u \right)} = \frac{e^{\frac{1}{u}}}{u^{2}}$$$:

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

Sea $$$v=\frac{1}{u}$$$.

Entonces $$$dv=\left(\frac{1}{u}\right)^{\prime }du = - \frac{1}{u^{2}} du$$$ (los pasos pueden verse »), y obtenemos que $$$\frac{du}{u^{2}} = - dv$$$.

Por lo tanto,

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

Aplica la regla del factor constante $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ con $$$c=-1$$$ y $$$f{\left(v \right)} = e^{v}$$$:

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

La integral de la función exponencial es $$$\int{e^{v} d v} = e^{v}$$$:

$$- \frac{{\color{red}{\int{e^{v} d v}}}}{3} = - \frac{{\color{red}{e^{v}}}}{3}$$

Recordemos que $$$v=\frac{1}{u}$$$:

$$- \frac{e^{{\color{red}{v}}}}{3} = - \frac{e^{{\color{red}{\frac{1}{u}}}}}{3}$$

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

$$- \frac{e^{{\color{red}{u}}^{-1}}}{3} = - \frac{e^{{\color{red}{x^{3}}}^{-1}}}{3}$$

Por lo tanto,

$$\int{\frac{e^{\frac{1}{x^{3}}}}{x^{4}} d x} = - \frac{e^{\frac{1}{x^{3}}}}{3}$$

Añade la constante de integración:

$$\int{\frac{e^{\frac{1}{x^{3}}}}{x^{4}} d x} = - \frac{e^{\frac{1}{x^{3}}}}{3}+C$$

$$$\int \frac{e^{\frac{1}{x^{3}}}}{x^{4}}\, dx = - \frac{e^{\frac{1}{x^{3}}}}{3} + C$$$A