Integral de $$$e^{\frac{x}{2}} - 2$$$

Encontre $$$\int \left(e^{\frac{x}{2}} - 2\right)\, dx$$$.

Integre termo a termo:

$${\color{red}{\int{\left(e^{\frac{x}{2}} - 2\right)d x}}} = {\color{red}{\left(- \int{2 d x} + \int{e^{\frac{x}{2}} d x}\right)}}$$

Aplique a regra da constante $$$\int c\, dx = c x$$$ usando $$$c=2$$$:

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

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

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

Portanto,

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

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

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

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

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

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

$$- 2 x + 2 e^{{\color{red}{u}}} = - 2 x + 2 e^{{\color{red}{\left(\frac{x}{2}\right)}}}$$

Portanto,

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

Adicione a constante de integração:

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

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