Integralen av $$$- 2 x - 2 - \frac{1}{x}$$$

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

Integrera termvis:

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

Tillämpa konstantregeln $$$\int c\, dx = c x$$$ med $$$c=2$$$:

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

Integralen av $$$\frac{1}{x}$$$ är $$$\int{\frac{1}{x} d x} = \ln{\left(\left|{x}\right| \right)}$$$:

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=2$$$ och $$$f{\left(x \right)} = x$$$:

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

Tillämpa potensregeln $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=1$$$:

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

Alltså,

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

Lägg till integrationskonstanten:

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

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