Integral of $$$2 x^{3} - \frac{1}{x^{3}}$$$

Find $$$\int \left(2 x^{3} - \frac{1}{x^{3}}\right)\, dx$$$.

Integrate term by term:

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

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

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

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=2$$$ and $$$f{\left(x \right)} = x^{3}$$$:

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

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

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

Therefore,

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

Simplify:

$$\int{\left(2 x^{3} - \frac{1}{x^{3}}\right)d x} = \frac{x^{6} + 1}{2 x^{2}}$$

Add the constant of integration:

$$\int{\left(2 x^{3} - \frac{1}{x^{3}}\right)d x} = \frac{x^{6} + 1}{2 x^{2}}+C$$

$$$\int \left(2 x^{3} - \frac{1}{x^{3}}\right)\, dx = \frac{x^{6} + 1}{2 x^{2}} + C$$$A