$$$\frac{1}{n^{4}}$$$'nin integrali

Bulun: $$$\int \frac{1}{n^{4}}\, dn$$$.

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

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

Dolayısıyla,

$$\int{\frac{1}{n^{4}} d n} = - \frac{1}{3 n^{3}}$$

İntegrasyon sabitini ekleyin:

$$\int{\frac{1}{n^{4}} d n} = - \frac{1}{3 n^{3}}+C$$

$$$\int \frac{1}{n^{4}}\, dn = - \frac{1}{3 n^{3}} + C$$$A