Integral de $$$e^{9 x}$$$

Encontre $$$\int e^{9 x}\, dx$$$.

Seja $$$u=9 x$$$.

Então $$$du=\left(9 x\right)^{\prime }dx = 9 dx$$$ (veja os passos »), e obtemos $$$dx = \frac{du}{9}$$$.

Logo,

$${\color{red}{\int{e^{9 x} d x}}} = {\color{red}{\int{\frac{e^{u}}{9} d u}}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=\frac{1}{9}$$$ e $$$f{\left(u \right)} = e^{u}$$$:

$${\color{red}{\int{\frac{e^{u}}{9} d u}}} = {\color{red}{\left(\frac{\int{e^{u} d u}}{9}\right)}}$$

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

$$\frac{{\color{red}{\int{e^{u} d u}}}}{9} = \frac{{\color{red}{e^{u}}}}{9}$$

Recorde que $$$u=9 x$$$:

$$\frac{e^{{\color{red}{u}}}}{9} = \frac{e^{{\color{red}{\left(9 x\right)}}}}{9}$$

Portanto,

$$\int{e^{9 x} d x} = \frac{e^{9 x}}{9}$$

Adicione a constante de integração:

$$\int{e^{9 x} d x} = \frac{e^{9 x}}{9}+C$$

$$$\int e^{9 x}\, dx = \frac{e^{9 x}}{9} + C$$$A