$$$2 x^{2} - 3 x - 2$$$ 的積分
您的輸入
求$$$\int \left(2 x^{2} - 3 x - 2\right)\, dx$$$。
解答
逐項積分:
$${\color{red}{\int{\left(2 x^{2} - 3 x - 2\right)d x}}} = {\color{red}{\left(- \int{2 d x} - \int{3 x d x} + \int{2 x^{2} d x}\right)}}$$
配合 $$$c=2$$$,應用常數法則 $$$\int c\, dx = c x$$$:
$$- \int{3 x d x} + \int{2 x^{2} d x} - {\color{red}{\int{2 d x}}} = - \int{3 x d x} + \int{2 x^{2} d x} - {\color{red}{\left(2 x\right)}}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$,使用 $$$c=3$$$ 與 $$$f{\left(x \right)} = x$$$:
$$- 2 x + \int{2 x^{2} d x} - {\color{red}{\int{3 x d x}}} = - 2 x + \int{2 x^{2} d x} - {\color{red}{\left(3 \int{x d x}\right)}}$$
套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,以 $$$n=1$$$:
$$- 2 x + \int{2 x^{2} d x} - 3 {\color{red}{\int{x d x}}}=- 2 x + \int{2 x^{2} d x} - 3 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=- 2 x + \int{2 x^{2} d x} - 3 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$
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 $$$f{\left(x \right)} = x^{2}$$$:
$$- \frac{3 x^{2}}{2} - 2 x + {\color{red}{\int{2 x^{2} d x}}} = - \frac{3 x^{2}}{2} - 2 x + {\color{red}{\left(2 \int{x^{2} d x}\right)}}$$
套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,以 $$$n=2$$$:
$$- \frac{3 x^{2}}{2} - 2 x + 2 {\color{red}{\int{x^{2} d x}}}=- \frac{3 x^{2}}{2} - 2 x + 2 {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=- \frac{3 x^{2}}{2} - 2 x + 2 {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$
因此,
$$\int{\left(2 x^{2} - 3 x - 2\right)d x} = \frac{2 x^{3}}{3} - \frac{3 x^{2}}{2} - 2 x$$
化簡:
$$\int{\left(2 x^{2} - 3 x - 2\right)d x} = \frac{x \left(4 x^{2} - 9 x - 12\right)}{6}$$
加上積分常數:
$$\int{\left(2 x^{2} - 3 x - 2\right)d x} = \frac{x \left(4 x^{2} - 9 x - 12\right)}{6}+C$$
答案
$$$\int \left(2 x^{2} - 3 x - 2\right)\, dx = \frac{x \left(4 x^{2} - 9 x - 12\right)}{6} + C$$$A