$$$1 + \frac{1}{x^{4}}$$$'nin integrali

Bulun: $$$\int \left(1 + \frac{1}{x^{4}}\right)\, dx$$$.

Her terimin integralini alın:

$${\color{red}{\int{\left(1 + \frac{1}{x^{4}}\right)d x}}} = {\color{red}{\left(\int{1 d x} + \int{\frac{1}{x^{4}} d x}\right)}}$$

$$$c=1$$$ kullanarak $$$\int c\, dx = c x$$$ sabit kuralını uygula:

$$\int{\frac{1}{x^{4}} d x} + {\color{red}{\int{1 d x}}} = \int{\frac{1}{x^{4}} d x} + {\color{red}{x}}$$

Kuvvet kuralını $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ $$$n=-4$$$ ile uygulayın:

$$x + {\color{red}{\int{\frac{1}{x^{4}} d x}}}=x + {\color{red}{\int{x^{-4} d x}}}=x + {\color{red}{\frac{x^{-4 + 1}}{-4 + 1}}}=x + {\color{red}{\left(- \frac{x^{-3}}{3}\right)}}=x + {\color{red}{\left(- \frac{1}{3 x^{3}}\right)}}$$

Dolayısıyla,

$$\int{\left(1 + \frac{1}{x^{4}}\right)d x} = x - \frac{1}{3 x^{3}}$$

İntegrasyon sabitini ekleyin:

$$\int{\left(1 + \frac{1}{x^{4}}\right)d x} = x - \frac{1}{3 x^{3}}+C$$

$$$\int \left(1 + \frac{1}{x^{4}}\right)\, dx = \left(x - \frac{1}{3 x^{3}}\right) + C$$$A