$$$- x + \sqrt{2} x$$$の積分

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

項別に積分せよ:

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

$$$n=1$$$ 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:

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

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

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

$$$n=1$$$ 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:

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

したがって、

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

簡単化せよ:

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

積分定数を加える:

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

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