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

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

Seja $$$u=x + 1$$$.

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

A integral pode ser reescrita como

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

Expand the expression:

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

Integre termo a termo:

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

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

$$\int{\frac{2}{u} d u} - {\color{red}{\int{1 d u}}} = \int{\frac{2}{u} d u} - {\color{red}{u}}$$

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

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

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

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

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

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

Portanto,

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

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

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

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