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

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

Let $$$u=x^{2} - 3$$$.

Then $$$du=\left(x^{2} - 3\right)^{\prime }dx = 2 x dx$$$ (steps can be seen »), and we have that $$$x dx = \frac{du}{2}$$$.

The integral can be rewritten as

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = u$$$:

$${\color{red}{\int{\frac{u}{2} d u}}} = {\color{red}{\left(\frac{\int{u d u}}{2}\right)}}$$

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

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

Recall that $$$u=x^{2} - 3$$$:

$$\frac{{\color{red}{u}}^{2}}{4} = \frac{{\color{red}{\left(x^{2} - 3\right)}}^{2}}{4}$$

Therefore,

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

Add the constant of integration:

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

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