Integral of $$$\frac{x^{2} \left(3 - \frac{1}{x^{2}}\right)}{3}$$$

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

The input is rewritten: $$$\int{\frac{x^{2} \left(3 - \frac{1}{x^{2}}\right)}{3} d x}=\int{x^{2} \left(1 - \frac{1}{3 x^{2}}\right) d x}$$$.

Expand the expression:

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=\frac{1}{3}$$$:

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

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

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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