Integral de $$$\frac{1}{\left(1 - x\right)^{2}}$$$

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

Seja $$$u=1 - x$$$.

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

A integral pode ser reescrita como

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

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

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

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

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

Recorde que $$$u=1 - x$$$:

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

Portanto,

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

Simplifique:

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

Adicione a constante de integração:

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

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