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

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

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

Então $$$du=\left(\sqrt{x}\right)^{\prime }dx = \frac{1}{2 \sqrt{x}} dx$$$ (veja os passos »), e obtemos $$$\frac{dx}{\sqrt{x}} = 2 du$$$.

Logo,

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

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=2$$$ e $$$f{\left(u \right)} = u e^{u}$$$:

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

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

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

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

Assim,

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

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

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

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

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

Portanto,

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

Simplifique:

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

Adicione a constante de integração:

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

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