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

Find $$$\int \frac{x - 1}{x^{5}}\, dx$$$.

Expand the expression:

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

Integrate term by term:

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

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-4$$$:

$$- \int{\frac{1}{x^{5}} d x} + {\color{red}{\int{\frac{1}{x^{4}} d x}}}=- \int{\frac{1}{x^{5}} d x} + {\color{red}{\int{x^{-4} d x}}}=- \int{\frac{1}{x^{5}} d x} + {\color{red}{\frac{x^{-4 + 1}}{-4 + 1}}}=- \int{\frac{1}{x^{5}} d x} + {\color{red}{\left(- \frac{x^{-3}}{3}\right)}}=- \int{\frac{1}{x^{5}} d x} + {\color{red}{\left(- \frac{1}{3 x^{3}}\right)}}$$

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

$$- {\color{red}{\int{\frac{1}{x^{5}} d x}}} - \frac{1}{3 x^{3}}=- {\color{red}{\int{x^{-5} d x}}} - \frac{1}{3 x^{3}}=- {\color{red}{\frac{x^{-5 + 1}}{-5 + 1}}} - \frac{1}{3 x^{3}}=- {\color{red}{\left(- \frac{x^{-4}}{4}\right)}} - \frac{1}{3 x^{3}}=- {\color{red}{\left(- \frac{1}{4 x^{4}}\right)}} - \frac{1}{3 x^{3}}$$

Therefore,

$$\int{\frac{x - 1}{x^{5}} d x} = - \frac{1}{3 x^{3}} + \frac{1}{4 x^{4}}$$

Simplify:

$$\int{\frac{x - 1}{x^{5}} d x} = \frac{3 - 4 x}{12 x^{4}}$$

Add the constant of integration:

$$\int{\frac{x - 1}{x^{5}} d x} = \frac{3 - 4 x}{12 x^{4}}+C$$

$$$\int \frac{x - 1}{x^{5}}\, dx = \frac{3 - 4 x}{12 x^{4}} + C$$$A