$$$62 x + \left(12 x - 12\right) e^{2} - 62$$$の積分

$$$\int \left(62 x + \left(12 x - 12\right) e^{2} - 62\right)\, dx$$$ を求めよ。

項別に積分せよ:

$${\color{red}{\int{\left(62 x + \left(12 x - 12\right) e^{2} - 62\right)d x}}} = {\color{red}{\left(- \int{62 d x} + \int{62 x d x} + \int{\left(12 x - 12\right) e^{2} d x}\right)}}$$

$$$c=62$$$ に対して定数則 $$$\int c\, dx = c x$$$ を適用する:

$$\int{62 x d x} + \int{\left(12 x - 12\right) e^{2} d x} - {\color{red}{\int{62 d x}}} = \int{62 x d x} + \int{\left(12 x - 12\right) e^{2} d x} - {\color{red}{\left(62 x\right)}}$$

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=62$$$ と $$$f{\left(x \right)} = x$$$ に対して適用する:

$$- 62 x + \int{\left(12 x - 12\right) e^{2} d x} + {\color{red}{\int{62 x d x}}} = - 62 x + \int{\left(12 x - 12\right) e^{2} d x} + {\color{red}{\left(62 \int{x d x}\right)}}$$

$$$n=1$$$ 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:

$$- 62 x + \int{\left(12 x - 12\right) e^{2} d x} + 62 {\color{red}{\int{x d x}}}=- 62 x + \int{\left(12 x - 12\right) e^{2} d x} + 62 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=- 62 x + \int{\left(12 x - 12\right) e^{2} d x} + 62 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

被積分関数を簡単化する:

$$31 x^{2} - 62 x + {\color{red}{\int{\left(12 x - 12\right) e^{2} d x}}} = 31 x^{2} - 62 x + {\color{red}{\int{12 \left(x - 1\right) e^{2} d x}}}$$

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=12 e^{2}$$$ と $$$f{\left(x \right)} = x - 1$$$ に対して適用する:

$$31 x^{2} - 62 x + {\color{red}{\int{12 \left(x - 1\right) e^{2} d x}}} = 31 x^{2} - 62 x + {\color{red}{\left(12 e^{2} \int{\left(x - 1\right)d x}\right)}}$$

項別に積分せよ:

$$31 x^{2} - 62 x + 12 e^{2} {\color{red}{\int{\left(x - 1\right)d x}}} = 31 x^{2} - 62 x + 12 e^{2} {\color{red}{\left(- \int{1 d x} + \int{x d x}\right)}}$$

$$$c=1$$$ に対して定数則 $$$\int c\, dx = c x$$$ を適用する:

$$31 x^{2} - 62 x + 12 e^{2} \left(\int{x d x} - {\color{red}{\int{1 d x}}}\right) = 31 x^{2} - 62 x + 12 e^{2} \left(\int{x d x} - {\color{red}{x}}\right)$$

$$$n=1$$$ 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:

$$31 x^{2} - 62 x + 12 e^{2} \left(- x + {\color{red}{\int{x d x}}}\right)=31 x^{2} - 62 x + 12 e^{2} \left(- x + {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}\right)=31 x^{2} - 62 x + 12 e^{2} \left(- x + {\color{red}{\left(\frac{x^{2}}{2}\right)}}\right)$$

したがって、

$$\int{\left(62 x + \left(12 x - 12\right) e^{2} - 62\right)d x} = 31 x^{2} - 62 x + 12 \left(\frac{x^{2}}{2} - x\right) e^{2}$$

簡単化せよ:

$$\int{\left(62 x + \left(12 x - 12\right) e^{2} - 62\right)d x} = x \left(31 + 6 e^{2}\right) \left(x - 2\right)$$

積分定数を加える:

$$\int{\left(62 x + \left(12 x - 12\right) e^{2} - 62\right)d x} = x \left(31 + 6 e^{2}\right) \left(x - 2\right)+C$$

$$$\int \left(62 x + \left(12 x - 12\right) e^{2} - 62\right)\, dx = x \left(31 + 6 e^{2}\right) \left(x - 2\right) + C$$$A