Integral de $$$\frac{2 - x}{1 - x}$$$

Encontre $$$\int \frac{2 - x}{1 - x}\, dx$$$.

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

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

Logo,

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

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

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

Expand the expression:

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

Integre termo a termo:

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

Aplique a regra da constante $$$\int c\, du = c u$$$ usando $$$c=1$$$:

$$- \int{\frac{1}{u} d u} - {\color{red}{\int{1 d u}}} = - \int{\frac{1}{u} d u} - {\color{red}{u}}$$

A integral de $$$\frac{1}{u}$$$ é $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

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

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

Portanto,

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

Adicione a constante de integração (e remova a constante da expressão):

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

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