$$$\left(3 x - 2\right) \left(4 x - 1\right)$$$の積分

$$$\int \left(3 x - 2\right) \left(4 x - 1\right)\, dx$$$ を求めよ。

Expand the expression:

$${\color{red}{\int{\left(3 x - 2\right) \left(4 x - 1\right) d x}}} = {\color{red}{\int{\left(12 x^{2} - 11 x + 2\right)d x}}}$$

項別に積分せよ:

$${\color{red}{\int{\left(12 x^{2} - 11 x + 2\right)d x}}} = {\color{red}{\left(\int{2 d x} - \int{11 x d x} + \int{12 x^{2} d x}\right)}}$$

$$$c=2$$$ に対して定数則 $$$\int c\, dx = c x$$$ を適用する:

$$- \int{11 x d x} + \int{12 x^{2} d x} + {\color{red}{\int{2 d x}}} = - \int{11 x d x} + \int{12 x^{2} d x} + {\color{red}{\left(2 x\right)}}$$

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=11$$$ と $$$f{\left(x \right)} = x$$$ に対して適用する:

$$2 x + \int{12 x^{2} d x} - {\color{red}{\int{11 x d x}}} = 2 x + \int{12 x^{2} d x} - {\color{red}{\left(11 \int{x d x}\right)}}$$

$$$n=1$$$ 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:

$$2 x + \int{12 x^{2} d x} - 11 {\color{red}{\int{x d x}}}=2 x + \int{12 x^{2} d x} - 11 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=2 x + \int{12 x^{2} d x} - 11 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=12$$$ と $$$f{\left(x \right)} = x^{2}$$$ に対して適用する:

$$- \frac{11 x^{2}}{2} + 2 x + {\color{red}{\int{12 x^{2} d x}}} = - \frac{11 x^{2}}{2} + 2 x + {\color{red}{\left(12 \int{x^{2} d x}\right)}}$$

$$$n=2$$$ 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:

$$- \frac{11 x^{2}}{2} + 2 x + 12 {\color{red}{\int{x^{2} d x}}}=- \frac{11 x^{2}}{2} + 2 x + 12 {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=- \frac{11 x^{2}}{2} + 2 x + 12 {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

したがって、

$$\int{\left(3 x - 2\right) \left(4 x - 1\right) d x} = 4 x^{3} - \frac{11 x^{2}}{2} + 2 x$$

簡単化せよ:

$$\int{\left(3 x - 2\right) \left(4 x - 1\right) d x} = \frac{x \left(8 x^{2} - 11 x + 4\right)}{2}$$

積分定数を加える:

$$\int{\left(3 x - 2\right) \left(4 x - 1\right) d x} = \frac{x \left(8 x^{2} - 11 x + 4\right)}{2}+C$$

$$$\int \left(3 x - 2\right) \left(4 x - 1\right)\, dx = \frac{x \left(8 x^{2} - 11 x + 4\right)}{2} + C$$$A