Integral de $$$1 - 6 e^{2 x}$$$

Encontre $$$\int \left(1 - 6 e^{2 x}\right)\, dx$$$.

Integre termo a termo:

$${\color{red}{\int{\left(1 - 6 e^{2 x}\right)d x}}} = {\color{red}{\left(\int{1 d x} - \int{6 e^{2 x} d x}\right)}}$$

Aplique a regra da constante $$$\int c\, dx = c x$$$ usando $$$c=1$$$:

$$- \int{6 e^{2 x} d x} + {\color{red}{\int{1 d x}}} = - \int{6 e^{2 x} d x} + {\color{red}{x}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=6$$$ e $$$f{\left(x \right)} = e^{2 x}$$$:

$$x - {\color{red}{\int{6 e^{2 x} d x}}} = x - {\color{red}{\left(6 \int{e^{2 x} d x}\right)}}$$

Seja $$$u=2 x$$$.

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

A integral torna-se

$$x - 6 {\color{red}{\int{e^{2 x} d x}}} = x - 6 {\color{red}{\int{\frac{e^{u}}{2} d u}}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=\frac{1}{2}$$$ e $$$f{\left(u \right)} = e^{u}$$$:

$$x - 6 {\color{red}{\int{\frac{e^{u}}{2} d u}}} = x - 6 {\color{red}{\left(\frac{\int{e^{u} d u}}{2}\right)}}$$

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

$$x - 3 {\color{red}{\int{e^{u} d u}}} = x - 3 {\color{red}{e^{u}}}$$

Recorde que $$$u=2 x$$$:

$$x - 3 e^{{\color{red}{u}}} = x - 3 e^{{\color{red}{\left(2 x\right)}}}$$

Portanto,

$$\int{\left(1 - 6 e^{2 x}\right)d x} = x - 3 e^{2 x}$$

Adicione a constante de integração:

$$\int{\left(1 - 6 e^{2 x}\right)d x} = x - 3 e^{2 x}+C$$

$$$\int \left(1 - 6 e^{2 x}\right)\, dx = \left(x - 3 e^{2 x}\right) + C$$$A