Integral of $$$\frac{x^{5} - 1}{x^{3}}$$$

Find $$$\int \frac{x^{5} - 1}{x^{3}}\, dx$$$.

Expand the expression:

$${\color{red}{\int{\frac{x^{5} - 1}{x^{3}} d x}}} = {\color{red}{\int{\left(x^{2} - \frac{1}{x^{3}}\right)d x}}}$$

Integrate term by term:

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

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:

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

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-3$$$:

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

Therefore,

$$\int{\frac{x^{5} - 1}{x^{3}} d x} = \frac{x^{3}}{3} + \frac{1}{2 x^{2}}$$

Simplify:

$$\int{\frac{x^{5} - 1}{x^{3}} d x} = \frac{2 x^{5} + 3}{6 x^{2}}$$

Add the constant of integration:

$$\int{\frac{x^{5} - 1}{x^{3}} d x} = \frac{2 x^{5} + 3}{6 x^{2}}+C$$

$$$\int \frac{x^{5} - 1}{x^{3}}\, dx = \frac{2 x^{5} + 3}{6 x^{2}} + C$$$A