Integral of $$$1 + \frac{1}{x^{5}}$$$

Find $$$\int \left(1 + \frac{1}{x^{5}}\right)\, dx$$$.

Integrate term by term:

$${\color{red}{\int{\left(1 + \frac{1}{x^{5}}\right)d x}}} = {\color{red}{\left(\int{1 d x} + \int{\frac{1}{x^{5}} d x}\right)}}$$

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:

$$\int{\frac{1}{x^{5}} d x} + {\color{red}{\int{1 d x}}} = \int{\frac{1}{x^{5}} d x} + {\color{red}{x}}$$

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-5$$$:

$$x + {\color{red}{\int{\frac{1}{x^{5}} d x}}}=x + {\color{red}{\int{x^{-5} d x}}}=x + {\color{red}{\frac{x^{-5 + 1}}{-5 + 1}}}=x + {\color{red}{\left(- \frac{x^{-4}}{4}\right)}}=x + {\color{red}{\left(- \frac{1}{4 x^{4}}\right)}}$$

Therefore,

$$\int{\left(1 + \frac{1}{x^{5}}\right)d x} = x - \frac{1}{4 x^{4}}$$

Add the constant of integration:

$$\int{\left(1 + \frac{1}{x^{5}}\right)d x} = x - \frac{1}{4 x^{4}}+C$$

$$$\int \left(1 + \frac{1}{x^{5}}\right)\, dx = \left(x - \frac{1}{4 x^{4}}\right) + C$$$A