Integral of $$$x^{4} - 6 x^{2}$$$
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Find $$$\int \left(x^{4} - 6 x^{2}\right)\, dx$$$.
Solution
Integrate term by term:
$${\color{red}{\int{\left(x^{4} - 6 x^{2}\right)d x}}} = {\color{red}{\left(- \int{6 x^{2} d x} + \int{x^{4} d x}\right)}}$$
Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=4$$$:
$$- \int{6 x^{2} d x} + {\color{red}{\int{x^{4} d x}}}=- \int{6 x^{2} d x} + {\color{red}{\frac{x^{1 + 4}}{1 + 4}}}=- \int{6 x^{2} d x} + {\color{red}{\left(\frac{x^{5}}{5}\right)}}$$
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=6$$$ and $$$f{\left(x \right)} = x^{2}$$$:
$$\frac{x^{5}}{5} - {\color{red}{\int{6 x^{2} d x}}} = \frac{x^{5}}{5} - {\color{red}{\left(6 \int{x^{2} d x}\right)}}$$
Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:
$$\frac{x^{5}}{5} - 6 {\color{red}{\int{x^{2} d x}}}=\frac{x^{5}}{5} - 6 {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=\frac{x^{5}}{5} - 6 {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$
Therefore,
$$\int{\left(x^{4} - 6 x^{2}\right)d x} = \frac{x^{5}}{5} - 2 x^{3}$$
Simplify:
$$\int{\left(x^{4} - 6 x^{2}\right)d x} = \frac{x^{3} \left(x^{2} - 10\right)}{5}$$
Add the constant of integration:
$$\int{\left(x^{4} - 6 x^{2}\right)d x} = \frac{x^{3} \left(x^{2} - 10\right)}{5}+C$$
Answer: $$$\int{\left(x^{4} - 6 x^{2}\right)d x}=\frac{x^{3} \left(x^{2} - 10\right)}{5}+C$$$