$$$x$$$ değişkenine göre $$$- \alpha \beta - x + \frac{1}{x}$$$ fonksiyonunun integrali

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

Her terimin integralini alın:

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

$$$\frac{1}{x}$$$'nin integrali $$$\int{\frac{1}{x} d x} = \ln{\left(\left|{x}\right| \right)}$$$:

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

Kuvvet kuralını $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ $$$n=1$$$ ile uygulayın:

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

$$$c=\alpha \beta$$$ kullanarak $$$\int c\, dx = c x$$$ sabit kuralını uygula:

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

Dolayısıyla,

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

İntegrasyon sabitini ekleyin:

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

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