Integral de $$$- \frac{3}{\left(x + 2\right)^{2}}$$$

Encontre $$$\int \left(- \frac{3}{\left(x + 2\right)^{2}}\right)\, dx$$$.

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=-3$$$ e $$$f{\left(x \right)} = \frac{1}{\left(x + 2\right)^{2}}$$$:

$${\color{red}{\int{\left(- \frac{3}{\left(x + 2\right)^{2}}\right)d x}}} = {\color{red}{\left(- 3 \int{\frac{1}{\left(x + 2\right)^{2}} d x}\right)}}$$

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

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

A integral torna-se

$$- 3 {\color{red}{\int{\frac{1}{\left(x + 2\right)^{2}} d x}}} = - 3 {\color{red}{\int{\frac{1}{u^{2}} d u}}}$$

Aplique a regra da potência $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=-2$$$:

$$- 3 {\color{red}{\int{\frac{1}{u^{2}} d u}}}=- 3 {\color{red}{\int{u^{-2} d u}}}=- 3 {\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}=- 3 {\color{red}{\left(- u^{-1}\right)}}=- 3 {\color{red}{\left(- \frac{1}{u}\right)}}$$

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

$$3 {\color{red}{u}}^{-1} = 3 {\color{red}{\left(x + 2\right)}}^{-1}$$

Portanto,

$$\int{\left(- \frac{3}{\left(x + 2\right)^{2}}\right)d x} = \frac{3}{x + 2}$$

Adicione a constante de integração:

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

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