Integral de $$$e^{\sqrt{x}}$$$

Halla $$$\int e^{\sqrt{x}}\, dx$$$.

Sea $$$u=\sqrt{x}$$$.

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

Entonces,

$${\color{red}{\int{e^{\sqrt{x}} d x}}} = {\color{red}{\int{2 u e^{u} d u}}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=2$$$ y $$$f{\left(u \right)} = u e^{u}$$$:

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

Para la integral $$$\int{u e^{u} d u}$$$, utiliza la integración por partes $$$\int \operatorname{m} \operatorname{dv} = \operatorname{m}\operatorname{v} - \int \operatorname{v} \operatorname{dm}$$$.

Sean $$$\operatorname{m}=u$$$ y $$$\operatorname{dv}=e^{u} du$$$.

Entonces $$$\operatorname{dm}=\left(u\right)^{\prime }du=1 du$$$ (los pasos pueden verse ») y $$$\operatorname{v}=\int{e^{u} d u}=e^{u}$$$ (los pasos pueden verse »).

La integral puede reescribirse como

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

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

$$2 u e^{u} - 2 {\color{red}{\int{e^{u} d u}}} = 2 u e^{u} - 2 {\color{red}{e^{u}}}$$

Recordemos que $$$u=\sqrt{x}$$$:

$$- 2 e^{{\color{red}{u}}} + 2 {\color{red}{u}} e^{{\color{red}{u}}} = - 2 e^{{\color{red}{\sqrt{x}}}} + 2 {\color{red}{\sqrt{x}}} e^{{\color{red}{\sqrt{x}}}}$$

Por lo tanto,

$$\int{e^{\sqrt{x}} d x} = 2 \sqrt{x} e^{\sqrt{x}} - 2 e^{\sqrt{x}}$$

Simplificar:

$$\int{e^{\sqrt{x}} d x} = 2 \left(\sqrt{x} - 1\right) e^{\sqrt{x}}$$

Añade la constante de integración:

$$\int{e^{\sqrt{x}} d x} = 2 \left(\sqrt{x} - 1\right) e^{\sqrt{x}}+C$$

$$$\int e^{\sqrt{x}}\, dx = 2 \left(\sqrt{x} - 1\right) e^{\sqrt{x}} + C$$$A