Integral of $$$- 4 x + \frac{1}{x^{3}}$$$

Find $$$\int \left(- 4 x + \frac{1}{x^{3}}\right)\, dx$$$.

Integrate term by term:

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

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

$$- \int{4 x d x} + {\color{red}{\int{\frac{1}{x^{3}} d x}}}=- \int{4 x d x} + {\color{red}{\int{x^{-3} d x}}}=- \int{4 x d x} + {\color{red}{\frac{x^{-3 + 1}}{-3 + 1}}}=- \int{4 x d x} + {\color{red}{\left(- \frac{x^{-2}}{2}\right)}}=- \int{4 x d x} + {\color{red}{\left(- \frac{1}{2 x^{2}}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=4$$$ and $$$f{\left(x \right)} = x$$$:

$$- {\color{red}{\int{4 x d x}}} - \frac{1}{2 x^{2}} = - {\color{red}{\left(4 \int{x d x}\right)}} - \frac{1}{2 x^{2}}$$

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

$$- 4 {\color{red}{\int{x d x}}} - \frac{1}{2 x^{2}}=- 4 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}} - \frac{1}{2 x^{2}}=- 4 {\color{red}{\left(\frac{x^{2}}{2}\right)}} - \frac{1}{2 x^{2}}$$

Therefore,

$$\int{\left(- 4 x + \frac{1}{x^{3}}\right)d x} = - 2 x^{2} - \frac{1}{2 x^{2}}$$

Simplify:

$$\int{\left(- 4 x + \frac{1}{x^{3}}\right)d x} = \frac{- 4 x^{4} - 1}{2 x^{2}}$$

Add the constant of integration:

$$\int{\left(- 4 x + \frac{1}{x^{3}}\right)d x} = \frac{- 4 x^{4} - 1}{2 x^{2}}+C$$

$$$\int \left(- 4 x + \frac{1}{x^{3}}\right)\, dx = \frac{- 4 x^{4} - 1}{2 x^{2}} + C$$$A