$$$- \frac{x}{e^{2}} + x e^{2}$$$の積分

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

項別に積分せよ:

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

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

$$- \int{\frac{x}{e^{2}} d x} + {\color{red}{\int{x e^{2} d x}}} = - \int{\frac{x}{e^{2}} d x} + {\color{red}{e^{2} \int{x d x}}}$$

$$$n=1$$$ 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:

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

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

$$\frac{x^{2} e^{2}}{2} - {\color{red}{\int{\frac{x}{e^{2}} d x}}} = \frac{x^{2} e^{2}}{2} - {\color{red}{\frac{\int{x d x}}{e^{2}}}}$$

$$$n=1$$$ 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:

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

したがって、

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

簡単化せよ:

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

積分定数を加える:

$$\int{\left(- \frac{x}{e^{2}} + x e^{2}\right)d x} = x^{2} \sinh{\left(2 \right)}+C$$

$$$\int \left(- \frac{x}{e^{2}} + x e^{2}\right)\, dx = x^{2} \sinh{\left(2 \right)} + C$$$A