$$$\frac{c}{x \left(c - x\right)}$$$ の $$$x$$$ に関する積分

$$$\int \frac{c}{x \left(c - x\right)}\, dx$$$ を求めよ。

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

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

部分分数分解を行う:

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

項別に積分せよ:

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

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

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

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

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

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

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

$$$u=c - x$$$ とする。

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

積分は次のようになります

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

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

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

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

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

次のことを思い出してください $$$u=c - x$$$:

$$c \left(\frac{\ln{\left(\left|{x}\right| \right)}}{c} - \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{c}\right) = c \left(\frac{\ln{\left(\left|{x}\right| \right)}}{c} - \frac{\ln{\left(\left|{{\color{red}{\left(c - x\right)}}}\right| \right)}}{c}\right)$$

したがって、

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

簡単化せよ:

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

積分定数を加える:

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

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