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

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

Integre termo a termo:

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

Aplique a regra da constante $$$\int c\, dx = c x$$$ usando $$$c=2$$$:

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

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

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

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

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

Portanto,

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

Adicione a constante de integração:

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

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