Integral of $$$x^{3} - 2$$$
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Find $$$\int \left(x^{3} - 2\right)\, dx$$$.
Solution
Integrate term by term:
$${\color{red}{\int{\left(x^{3} - 2\right)d x}}} = {\color{red}{\left(- \int{2 d x} + \int{x^{3} d x}\right)}}$$
Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=2$$$:
$$\int{x^{3} d x} - {\color{red}{\int{2 d x}}} = \int{x^{3} d x} - {\color{red}{\left(2 x\right)}}$$
Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=3$$$:
$$- 2 x + {\color{red}{\int{x^{3} d x}}}=- 2 x + {\color{red}{\frac{x^{1 + 3}}{1 + 3}}}=- 2 x + {\color{red}{\left(\frac{x^{4}}{4}\right)}}$$
Therefore,
$$\int{\left(x^{3} - 2\right)d x} = \frac{x^{4}}{4} - 2 x$$
Simplify:
$$\int{\left(x^{3} - 2\right)d x} = \frac{x \left(x^{3} - 8\right)}{4}$$
Add the constant of integration:
$$\int{\left(x^{3} - 2\right)d x} = \frac{x \left(x^{3} - 8\right)}{4}+C$$
Answer
$$$\int \left(x^{3} - 2\right)\, dx = \frac{x \left(x^{3} - 8\right)}{4} + C$$$A