Integral de $$$e^{\sqrt[3]{x}}$$$
Calculadora relacionada: Calculadora de integrales definidas e impropias
Tu entrada
Halla $$$\int e^{\sqrt[3]{x}}\, dx$$$.
Solución
Sea $$$u=\sqrt[3]{x}$$$.
Entonces $$$du=\left(\sqrt[3]{x}\right)^{\prime }dx = \frac{1}{3 x^{\frac{2}{3}}} dx$$$ (los pasos pueden verse »), y obtenemos que $$$\frac{dx}{x^{\frac{2}{3}}} = 3 du$$$.
Entonces,
$${\color{red}{\int{e^{\sqrt[3]{x}} d x}}} = {\color{red}{\int{3 u^{2} e^{u} d u}}}$$
Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=3$$$ y $$$f{\left(u \right)} = u^{2} e^{u}$$$:
$${\color{red}{\int{3 u^{2} e^{u} d u}}} = {\color{red}{\left(3 \int{u^{2} e^{u} d u}\right)}}$$
Para la integral $$$\int{u^{2} e^{u} d u}$$$, utiliza la integración por partes $$$\int \operatorname{\mu} \operatorname{dv} = \operatorname{\mu}\operatorname{v} - \int \operatorname{v} \operatorname{d\mu}$$$.
Sean $$$\operatorname{\mu}=u^{2}$$$ y $$$\operatorname{dv}=e^{u} du$$$.
Entonces $$$\operatorname{d\mu}=\left(u^{2}\right)^{\prime }du=2 u du$$$ (los pasos pueden verse ») y $$$\operatorname{v}=\int{e^{u} d u}=e^{u}$$$ (los pasos pueden verse »).
La integral se convierte en
$$3 {\color{red}{\int{u^{2} e^{u} d u}}}=3 {\color{red}{\left(u^{2} \cdot e^{u}-\int{e^{u} \cdot 2 u d u}\right)}}=3 {\color{red}{\left(u^{2} e^{u} - \int{2 u e^{u} d u}\right)}}$$
Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=2$$$ y $$$f{\left(u \right)} = u e^{u}$$$:
$$3 u^{2} e^{u} - 3 {\color{red}{\int{2 u e^{u} d u}}} = 3 u^{2} e^{u} - 3 {\color{red}{\left(2 \int{u e^{u} d u}\right)}}$$
Para la integral $$$\int{u e^{u} d u}$$$, utiliza la integración por partes $$$\int \operatorname{\mu} \operatorname{dv} = \operatorname{\mu}\operatorname{v} - \int \operatorname{v} \operatorname{d\mu}$$$.
Sean $$$\operatorname{\mu}=u$$$ y $$$\operatorname{dv}=e^{u} du$$$.
Entonces $$$\operatorname{d\mu}=\left(u\right)^{\prime }du=1 du$$$ (los pasos pueden verse ») y $$$\operatorname{v}=\int{e^{u} d u}=e^{u}$$$ (los pasos pueden verse »).
La integral se convierte en
$$3 u^{2} e^{u} - 6 {\color{red}{\int{u e^{u} d u}}}=3 u^{2} e^{u} - 6 {\color{red}{\left(u \cdot e^{u}-\int{e^{u} \cdot 1 d u}\right)}}=3 u^{2} e^{u} - 6 {\color{red}{\left(u e^{u} - \int{e^{u} d u}\right)}}$$
La integral de la función exponencial es $$$\int{e^{u} d u} = e^{u}$$$:
$$3 u^{2} e^{u} - 6 u e^{u} + 6 {\color{red}{\int{e^{u} d u}}} = 3 u^{2} e^{u} - 6 u e^{u} + 6 {\color{red}{e^{u}}}$$
Recordemos que $$$u=\sqrt[3]{x}$$$:
$$6 e^{{\color{red}{u}}} - 6 {\color{red}{u}} e^{{\color{red}{u}}} + 3 {\color{red}{u}}^{2} e^{{\color{red}{u}}} = 6 e^{{\color{red}{\sqrt[3]{x}}}} - 6 {\color{red}{\sqrt[3]{x}}} e^{{\color{red}{\sqrt[3]{x}}}} + 3 {\color{red}{\sqrt[3]{x}}}^{2} e^{{\color{red}{\sqrt[3]{x}}}}$$
Por lo tanto,
$$\int{e^{\sqrt[3]{x}} d x} = 3 x^{\frac{2}{3}} e^{\sqrt[3]{x}} - 6 \sqrt[3]{x} e^{\sqrt[3]{x}} + 6 e^{\sqrt[3]{x}}$$
Simplificar:
$$\int{e^{\sqrt[3]{x}} d x} = 3 \left(x^{\frac{2}{3}} - 2 \sqrt[3]{x} + 2\right) e^{\sqrt[3]{x}}$$
Añade la constante de integración:
$$\int{e^{\sqrt[3]{x}} d x} = 3 \left(x^{\frac{2}{3}} - 2 \sqrt[3]{x} + 2\right) e^{\sqrt[3]{x}}+C$$
Respuesta
$$$\int e^{\sqrt[3]{x}}\, dx = 3 \left(x^{\frac{2}{3}} - 2 \sqrt[3]{x} + 2\right) e^{\sqrt[3]{x}} + C$$$A