$$$\frac{4 \ln\left(2 x\right) - 9}{x}$$$の積分

$$$\int \frac{4 \ln\left(2 x\right) - 9}{x}\, dx$$$ を求めよ。

Expand the expression:

$${\color{red}{\int{\frac{4 \ln{\left(2 x \right)} - 9}{x} d x}}} = {\color{red}{\int{\left(\frac{4 \ln{\left(x \right)}}{x} - \frac{9}{x} + \frac{4 \ln{\left(2 \right)}}{x}\right)d x}}}$$

項別に積分せよ:

$${\color{red}{\int{\left(\frac{4 \ln{\left(x \right)}}{x} - \frac{9}{x} + \frac{4 \ln{\left(2 \right)}}{x}\right)d x}}} = {\color{red}{\left(- \int{\frac{9}{x} d x} + \int{\frac{4 \ln{\left(2 \right)}}{x} d x} + \int{\frac{4 \ln{\left(x \right)}}{x} d x}\right)}}$$

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

$$\int{\frac{4 \ln{\left(2 \right)}}{x} d x} + \int{\frac{4 \ln{\left(x \right)}}{x} d x} - {\color{red}{\int{\frac{9}{x} d x}}} = \int{\frac{4 \ln{\left(2 \right)}}{x} d x} + \int{\frac{4 \ln{\left(x \right)}}{x} d x} - {\color{red}{\left(9 \int{\frac{1}{x} d x}\right)}}$$

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

$$\int{\frac{4 \ln{\left(2 \right)}}{x} d x} + \int{\frac{4 \ln{\left(x \right)}}{x} d x} - 9 {\color{red}{\int{\frac{1}{x} d x}}} = \int{\frac{4 \ln{\left(2 \right)}}{x} d x} + \int{\frac{4 \ln{\left(x \right)}}{x} d x} - 9 {\color{red}{\ln{\left(\left|{x}\right| \right)}}}$$

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=4 \ln{\left(2 \right)}$$$ と $$$f{\left(x \right)} = \frac{1}{x}$$$ に対して適用する:

$$- 9 \ln{\left(\left|{x}\right| \right)} + \int{\frac{4 \ln{\left(x \right)}}{x} d x} + {\color{red}{\int{\frac{4 \ln{\left(2 \right)}}{x} d x}}} = - 9 \ln{\left(\left|{x}\right| \right)} + \int{\frac{4 \ln{\left(x \right)}}{x} d x} + {\color{red}{\left(4 \ln{\left(2 \right)} \int{\frac{1}{x} d x}\right)}}$$

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

$$- 9 \ln{\left(\left|{x}\right| \right)} + \int{\frac{4 \ln{\left(x \right)}}{x} d x} + 4 \ln{\left(2 \right)} {\color{red}{\int{\frac{1}{x} d x}}} = - 9 \ln{\left(\left|{x}\right| \right)} + \int{\frac{4 \ln{\left(x \right)}}{x} d x} + 4 \ln{\left(2 \right)} {\color{red}{\ln{\left(\left|{x}\right| \right)}}}$$

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

$$- 9 \ln{\left(\left|{x}\right| \right)} + 4 \ln{\left(2 \right)} \ln{\left(\left|{x}\right| \right)} + {\color{red}{\int{\frac{4 \ln{\left(x \right)}}{x} d x}}} = - 9 \ln{\left(\left|{x}\right| \right)} + 4 \ln{\left(2 \right)} \ln{\left(\left|{x}\right| \right)} + {\color{red}{\left(4 \int{\frac{\ln{\left(x \right)}}{x} d x}\right)}}$$

$$$u=\ln{\left(x \right)}$$$ とする。

すると $$$du=\left(\ln{\left(x \right)}\right)^{\prime }dx = \frac{dx}{x}$$$（手順は»で確認できます）、$$$\frac{dx}{x} = du$$$ となります。

したがって、

$$- 9 \ln{\left(\left|{x}\right| \right)} + 4 \ln{\left(2 \right)} \ln{\left(\left|{x}\right| \right)} + 4 {\color{red}{\int{\frac{\ln{\left(x \right)}}{x} d x}}} = - 9 \ln{\left(\left|{x}\right| \right)} + 4 \ln{\left(2 \right)} \ln{\left(\left|{x}\right| \right)} + 4 {\color{red}{\int{u d u}}}$$

$$$n=1$$$ を用いて、べき乗の法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ を適用します:

$$- 9 \ln{\left(\left|{x}\right| \right)} + 4 \ln{\left(2 \right)} \ln{\left(\left|{x}\right| \right)} + 4 {\color{red}{\int{u d u}}}=- 9 \ln{\left(\left|{x}\right| \right)} + 4 \ln{\left(2 \right)} \ln{\left(\left|{x}\right| \right)} + 4 {\color{red}{\frac{u^{1 + 1}}{1 + 1}}}=- 9 \ln{\left(\left|{x}\right| \right)} + 4 \ln{\left(2 \right)} \ln{\left(\left|{x}\right| \right)} + 4 {\color{red}{\left(\frac{u^{2}}{2}\right)}}$$

次のことを思い出してください $$$u=\ln{\left(x \right)}$$$:

$$- 9 \ln{\left(\left|{x}\right| \right)} + 4 \ln{\left(2 \right)} \ln{\left(\left|{x}\right| \right)} + 2 {\color{red}{u}}^{2} = - 9 \ln{\left(\left|{x}\right| \right)} + 4 \ln{\left(2 \right)} \ln{\left(\left|{x}\right| \right)} + 2 {\color{red}{\ln{\left(x \right)}}}^{2}$$

したがって、

$$\int{\frac{4 \ln{\left(2 x \right)} - 9}{x} d x} = 2 \ln{\left(x \right)}^{2} - 9 \ln{\left(\left|{x}\right| \right)} + 4 \ln{\left(2 \right)} \ln{\left(\left|{x}\right| \right)}$$

積分定数を加える:

$$\int{\frac{4 \ln{\left(2 x \right)} - 9}{x} d x} = 2 \ln{\left(x \right)}^{2} - 9 \ln{\left(\left|{x}\right| \right)} + 4 \ln{\left(2 \right)} \ln{\left(\left|{x}\right| \right)}+C$$

$$$\int \frac{4 \ln\left(2 x\right) - 9}{x}\, dx = \left(2 \ln^{2}\left(x\right) - 9 \ln\left(\left|{x}\right|\right) + 4 \ln\left(2\right) \ln\left(\left|{x}\right|\right)\right) + C$$$A