$$$2 x^{2} \left(2 x - 4\right)$$$ 的積分
您的輸入
求$$$\int 2 x^{2} \left(2 x - 4\right)\, dx$$$。
解答
已將輸入重寫為：$$$\int{2 x^{2} \left(2 x - 4\right) d x}=\int{x^{2} \left(4 x - 8\right) d x}$$$。
簡化被積函數:
$${\color{red}{\int{x^{2} \left(4 x - 8\right) d x}}} = {\color{red}{\int{4 x^{2} \left(x - 2\right) d x}}}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=4$$$ 與 $$$f{\left(x \right)} = x^{2} \left(x - 2\right)$$$：
$${\color{red}{\int{4 x^{2} \left(x - 2\right) d x}}} = {\color{red}{\left(4 \int{x^{2} \left(x - 2\right) d x}\right)}}$$
Expand the expression:
$$4 {\color{red}{\int{x^{2} \left(x - 2\right) d x}}} = 4 {\color{red}{\int{\left(x^{3} - 2 x^{2}\right)d x}}}$$
逐項積分：
$$4 {\color{red}{\int{\left(x^{3} - 2 x^{2}\right)d x}}} = 4 {\color{red}{\left(- \int{2 x^{2} d x} + \int{x^{3} d x}\right)}}$$
套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=3$$$：
$$- 4 \int{2 x^{2} d x} + 4 {\color{red}{\int{x^{3} d x}}}=- 4 \int{2 x^{2} d x} + 4 {\color{red}{\frac{x^{1 + 3}}{1 + 3}}}=- 4 \int{2 x^{2} d x} + 4 {\color{red}{\left(\frac{x^{4}}{4}\right)}}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=2$$$ 與 $$$f{\left(x \right)} = x^{2}$$$：
$$x^{4} - 4 {\color{red}{\int{2 x^{2} d x}}} = x^{4} - 4 {\color{red}{\left(2 \int{x^{2} d x}\right)}}$$
套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=2$$$：
$$x^{4} - 8 {\color{red}{\int{x^{2} d x}}}=x^{4} - 8 {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=x^{4} - 8 {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$
因此，
$$\int{x^{2} \left(4 x - 8\right) d x} = x^{4} - \frac{8 x^{3}}{3}$$
化簡：
$$\int{x^{2} \left(4 x - 8\right) d x} = x^{3} \left(x - \frac{8}{3}\right)$$
加上積分常數：
$$\int{x^{2} \left(4 x - 8\right) d x} = x^{3} \left(x - \frac{8}{3}\right)+C$$
答案
$$$\int 2 x^{2} \left(2 x - 4\right)\, dx = x^{3} \left(x - \frac{8}{3}\right) + C$$$A