Integral de $$$e^{\sqrt{2} \sqrt{x}}$$$

Halla $$$\int e^{\sqrt{2} \sqrt{x}}\, dx$$$.

Sea $$$u=\sqrt{2} \sqrt{x}$$$.

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

Entonces,

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

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

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

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

Por lo tanto,

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

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

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

Recordemos que $$$u=\sqrt{2} \sqrt{x}$$$:

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

Por lo tanto,

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

Simplificar:

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

Añade la constante de integración:

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

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