$$$\frac{1}{x} - \frac{1}{3 x^{3}}$$$'nin integrali

Bulun: $$$\int \left(\frac{1}{x} - \frac{1}{3 x^{3}}\right)\, dx$$$.

Her terimin integralini alın:

$${\color{red}{\int{\left(\frac{1}{x} - \frac{1}{3 x^{3}}\right)d x}}} = {\color{red}{\left(- \int{\frac{1}{3 x^{3}} d x} + \int{\frac{1}{x} d x}\right)}}$$

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

$$- \int{\frac{1}{3 x^{3}} d x} + {\color{red}{\int{\frac{1}{x} d x}}} = - \int{\frac{1}{3 x^{3}} d x} + {\color{red}{\ln{\left(\left|{x}\right| \right)}}}$$

Sabit katsayı kuralı $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$'i $$$c=\frac{1}{3}$$$ ve $$$f{\left(x \right)} = \frac{1}{x^{3}}$$$ ile uygula:

$$\ln{\left(\left|{x}\right| \right)} - {\color{red}{\int{\frac{1}{3 x^{3}} d x}}} = \ln{\left(\left|{x}\right| \right)} - {\color{red}{\left(\frac{\int{\frac{1}{x^{3}} d x}}{3}\right)}}$$

Kuvvet kuralını $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ $$$n=-3$$$ ile uygulayın:

$$\ln{\left(\left|{x}\right| \right)} - \frac{{\color{red}{\int{\frac{1}{x^{3}} d x}}}}{3}=\ln{\left(\left|{x}\right| \right)} - \frac{{\color{red}{\int{x^{-3} d x}}}}{3}=\ln{\left(\left|{x}\right| \right)} - \frac{{\color{red}{\frac{x^{-3 + 1}}{-3 + 1}}}}{3}=\ln{\left(\left|{x}\right| \right)} - \frac{{\color{red}{\left(- \frac{x^{-2}}{2}\right)}}}{3}=\ln{\left(\left|{x}\right| \right)} - \frac{{\color{red}{\left(- \frac{1}{2 x^{2}}\right)}}}{3}$$

Dolayısıyla,

$$\int{\left(\frac{1}{x} - \frac{1}{3 x^{3}}\right)d x} = \ln{\left(\left|{x}\right| \right)} + \frac{1}{6 x^{2}}$$

İntegrasyon sabitini ekleyin:

$$\int{\left(\frac{1}{x} - \frac{1}{3 x^{3}}\right)d x} = \ln{\left(\left|{x}\right| \right)} + \frac{1}{6 x^{2}}+C$$

$$$\int \left(\frac{1}{x} - \frac{1}{3 x^{3}}\right)\, dx = \left(\ln\left(\left|{x}\right|\right) + \frac{1}{6 x^{2}}\right) + C$$$A