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

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

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

Então $$$du=\left(\sqrt{2} \sqrt{x}\right)^{\prime }dx = \frac{\sqrt{2}}{2 \sqrt{x}} dx$$$ (veja os passos »), e obtemos $$$\frac{dx}{\sqrt{x}} = \sqrt{2} du$$$.

Portanto,

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

Para a integral $$$\int{u e^{u} d u}$$$, use integração por partes $$$\int \operatorname{g} \operatorname{dv} = \operatorname{g}\operatorname{v} - \int \operatorname{v} \operatorname{dg}$$$.

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

Então $$$\operatorname{dg}=\left(u\right)^{\prime }du=1 du$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{e^{u} d u}=e^{u}$$$ (os passos podem ser vistos »).

Logo,

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

A integral da função exponencial é $$$\int{e^{u} d u} = e^{u}$$$:

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

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

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

Portanto,

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

Simplifique:

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

Adicione a constante de integração:

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

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