Integralen av $$$- \alpha \beta - x + \frac{1}{x}$$$ med avseende på $$$x$$$

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

Integrera termvis:

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

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

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

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

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

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

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

Alltså,

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

Lägg till integrationskonstanten:

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

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