$$$x^{4} - 6 x^{2}$$$の積分

$$$\int \left(x^{4} - 6 x^{2}\right)\, dx$$$ を求めよ。

項別に積分せよ:

$${\color{red}{\int{\left(x^{4} - 6 x^{2}\right)d x}}} = {\color{red}{\left(- \int{6 x^{2} d x} + \int{x^{4} d x}\right)}}$$

$$$n=4$$$ を用いて、べき乗の法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ を適用します:

$$- \int{6 x^{2} d x} + {\color{red}{\int{x^{4} d x}}}=- \int{6 x^{2} d x} + {\color{red}{\frac{x^{1 + 4}}{1 + 4}}}=- \int{6 x^{2} d x} + {\color{red}{\left(\frac{x^{5}}{5}\right)}}$$

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=6$$$ と $$$f{\left(x \right)} = x^{2}$$$ に対して適用する:

$$\frac{x^{5}}{5} - {\color{red}{\int{6 x^{2} d x}}} = \frac{x^{5}}{5} - {\color{red}{\left(6 \int{x^{2} d x}\right)}}$$

$$$n=2$$$ を用いて、べき乗の法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ を適用します:

$$\frac{x^{5}}{5} - 6 {\color{red}{\int{x^{2} d x}}}=\frac{x^{5}}{5} - 6 {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=\frac{x^{5}}{5} - 6 {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

したがって、

$$\int{\left(x^{4} - 6 x^{2}\right)d x} = \frac{x^{5}}{5} - 2 x^{3}$$

簡単化せよ:

$$\int{\left(x^{4} - 6 x^{2}\right)d x} = \frac{x^{3} \left(x^{2} - 10\right)}{5}$$

積分定数を加える:

$$\int{\left(x^{4} - 6 x^{2}\right)d x} = \frac{x^{3} \left(x^{2} - 10\right)}{5}+C$$

$$$\int \left(x^{4} - 6 x^{2}\right)\, dx = \frac{x^{3} \left(x^{2} - 10\right)}{5} + C$$$A