$$$- x + \ln\left(2\right)$$$の積分

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

項別に積分せよ:

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

$$$n=1$$$ 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:

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

$$$c=\ln{\left(2 \right)}$$$ に対して定数則 $$$\int c\, dx = c x$$$ を適用する:

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

したがって、

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

簡単化せよ:

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

積分定数を加える:

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

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