$$$- 4 x + \frac{\sqrt{5} x}{5}$$$の積分

$$$\int \left(- 4 x + \frac{\sqrt{5} x}{5}\right)\, dx$$$ を求めよ。

項別に積分せよ:

$${\color{red}{\int{\left(- 4 x + \frac{\sqrt{5} x}{5}\right)d x}}} = {\color{red}{\left(- \int{4 x d x} + \int{\frac{\sqrt{5} x}{5} d x}\right)}}$$

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=4$$$ と $$$f{\left(x \right)} = x$$$ に対して適用する:

$$\int{\frac{\sqrt{5} x}{5} d x} - {\color{red}{\int{4 x d x}}} = \int{\frac{\sqrt{5} x}{5} d x} - {\color{red}{\left(4 \int{x d x}\right)}}$$

$$$n=1$$$ 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:

$$\int{\frac{\sqrt{5} x}{5} d x} - 4 {\color{red}{\int{x d x}}}=\int{\frac{\sqrt{5} x}{5} d x} - 4 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=\int{\frac{\sqrt{5} x}{5} d x} - 4 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=\frac{\sqrt{5}}{5}$$$ と $$$f{\left(x \right)} = x$$$ に対して適用する:

$$- 2 x^{2} + {\color{red}{\int{\frac{\sqrt{5} x}{5} d x}}} = - 2 x^{2} + {\color{red}{\left(\frac{\sqrt{5} \int{x d x}}{5}\right)}}$$

$$$n=1$$$ 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:

$$- 2 x^{2} + \frac{\sqrt{5} {\color{red}{\int{x d x}}}}{5}=- 2 x^{2} + \frac{\sqrt{5} {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}}{5}=- 2 x^{2} + \frac{\sqrt{5} {\color{red}{\left(\frac{x^{2}}{2}\right)}}}{5}$$

したがって、

$$\int{\left(- 4 x + \frac{\sqrt{5} x}{5}\right)d x} = - 2 x^{2} + \frac{\sqrt{5} x^{2}}{10}$$

簡単化せよ:

$$\int{\left(- 4 x + \frac{\sqrt{5} x}{5}\right)d x} = \frac{x^{2} \left(-20 + \sqrt{5}\right)}{10}$$

積分定数を加える:

$$\int{\left(- 4 x + \frac{\sqrt{5} x}{5}\right)d x} = \frac{x^{2} \left(-20 + \sqrt{5}\right)}{10}+C$$

$$$\int \left(- 4 x + \frac{\sqrt{5} x}{5}\right)\, dx = \frac{x^{2} \left(-20 + \sqrt{5}\right)}{10} + C$$$A