$$$\frac{16 x - 9}{x}$$$の積分

$$$\int \frac{16 x - 9}{x}\, dx$$$ を求めよ。

Expand the expression:

$${\color{red}{\int{\frac{16 x - 9}{x} d x}}} = {\color{red}{\int{\left(16 - \frac{9}{x}\right)d x}}}$$

項別に積分せよ:

$${\color{red}{\int{\left(16 - \frac{9}{x}\right)d x}}} = {\color{red}{\left(\int{16 d x} - \int{\frac{9}{x} d x}\right)}}$$

$$$c=16$$$ に対して定数則 $$$\int c\, dx = c x$$$ を適用する:

$$- \int{\frac{9}{x} d x} + {\color{red}{\int{16 d x}}} = - \int{\frac{9}{x} d x} + {\color{red}{\left(16 x\right)}}$$

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=9$$$ と $$$f{\left(x \right)} = \frac{1}{x}$$$ に対して適用する:

$$16 x - {\color{red}{\int{\frac{9}{x} d x}}} = 16 x - {\color{red}{\left(9 \int{\frac{1}{x} d x}\right)}}$$

$$$\frac{1}{x}$$$ の不定積分は $$$\int{\frac{1}{x} d x} = \ln{\left(\left|{x}\right| \right)}$$$ です:

$$16 x - 9 {\color{red}{\int{\frac{1}{x} d x}}} = 16 x - 9 {\color{red}{\ln{\left(\left|{x}\right| \right)}}}$$

したがって、

$$\int{\frac{16 x - 9}{x} d x} = 16 x - 9 \ln{\left(\left|{x}\right| \right)}$$

積分定数を加える:

$$\int{\frac{16 x - 9}{x} d x} = 16 x - 9 \ln{\left(\left|{x}\right| \right)}+C$$

$$$\int \frac{16 x - 9}{x}\, dx = \left(16 x - 9 \ln\left(\left|{x}\right|\right)\right) + C$$$A