$$$\frac{1}{5 x^{3} - 5 x^{2}}$$$の積分

$$$\int \frac{1}{5 x^{3} - 5 x^{2}}\, dx$$$ を求めよ。

被積分関数を簡単化する:

$${\color{red}{\int{\frac{1}{5 x^{3} - 5 x^{2}} d x}}} = {\color{red}{\int{\frac{1}{5 x^{2} \left(x - 1\right)} d x}}}$$

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=\frac{1}{5}$$$ と $$$f{\left(x \right)} = \frac{1}{x^{2} \left(x - 1\right)}$$$ に対して適用する:

$${\color{red}{\int{\frac{1}{5 x^{2} \left(x - 1\right)} d x}}} = {\color{red}{\left(\frac{\int{\frac{1}{x^{2} \left(x - 1\right)} d x}}{5}\right)}}$$

部分分数分解を行う (手順は»で確認できます):

$$\frac{{\color{red}{\int{\frac{1}{x^{2} \left(x - 1\right)} d x}}}}{5} = \frac{{\color{red}{\int{\left(\frac{1}{x - 1} - \frac{1}{x} - \frac{1}{x^{2}}\right)d x}}}}{5}$$

項別に積分せよ:

$$\frac{{\color{red}{\int{\left(\frac{1}{x - 1} - \frac{1}{x} - \frac{1}{x^{2}}\right)d x}}}}{5} = \frac{{\color{red}{\left(- \int{\frac{1}{x^{2}} d x} - \int{\frac{1}{x} d x} + \int{\frac{1}{x - 1} d x}\right)}}}{5}$$

$$$u=x - 1$$$ とする。

すると $$$du=\left(x - 1\right)^{\prime }dx = 1 dx$$$（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

$$- \frac{\int{\frac{1}{x^{2}} d x}}{5} - \frac{\int{\frac{1}{x} d x}}{5} + \frac{{\color{red}{\int{\frac{1}{x - 1} d x}}}}{5} = - \frac{\int{\frac{1}{x^{2}} d x}}{5} - \frac{\int{\frac{1}{x} d x}}{5} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{5}$$

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

$$- \frac{\int{\frac{1}{x^{2}} d x}}{5} - \frac{\int{\frac{1}{x} d x}}{5} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{5} = - \frac{\int{\frac{1}{x^{2}} d x}}{5} - \frac{\int{\frac{1}{x} d x}}{5} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{5}$$

次のことを思い出してください $$$u=x - 1$$$:

$$\frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{5} - \frac{\int{\frac{1}{x^{2}} d x}}{5} - \frac{\int{\frac{1}{x} d x}}{5} = \frac{\ln{\left(\left|{{\color{red}{\left(x - 1\right)}}}\right| \right)}}{5} - \frac{\int{\frac{1}{x^{2}} d x}}{5} - \frac{\int{\frac{1}{x} d x}}{5}$$

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

$$\frac{\ln{\left(\left|{x - 1}\right| \right)}}{5} - \frac{\int{\frac{1}{x^{2}} d x}}{5} - \frac{{\color{red}{\int{\frac{1}{x} d x}}}}{5} = \frac{\ln{\left(\left|{x - 1}\right| \right)}}{5} - \frac{\int{\frac{1}{x^{2}} d x}}{5} - \frac{{\color{red}{\ln{\left(\left|{x}\right| \right)}}}}{5}$$

$$$n=-2$$$ 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:

$$- \frac{\ln{\left(\left|{x}\right| \right)}}{5} + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{5} - \frac{{\color{red}{\int{\frac{1}{x^{2}} d x}}}}{5}=- \frac{\ln{\left(\left|{x}\right| \right)}}{5} + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{5} - \frac{{\color{red}{\int{x^{-2} d x}}}}{5}=- \frac{\ln{\left(\left|{x}\right| \right)}}{5} + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{5} - \frac{{\color{red}{\frac{x^{-2 + 1}}{-2 + 1}}}}{5}=- \frac{\ln{\left(\left|{x}\right| \right)}}{5} + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{5} - \frac{{\color{red}{\left(- x^{-1}\right)}}}{5}=- \frac{\ln{\left(\left|{x}\right| \right)}}{5} + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{5} - \frac{{\color{red}{\left(- \frac{1}{x}\right)}}}{5}$$

したがって、

$$\int{\frac{1}{5 x^{3} - 5 x^{2}} d x} = - \frac{\ln{\left(\left|{x}\right| \right)}}{5} + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{5} + \frac{1}{5 x}$$

簡単化せよ:

$$\int{\frac{1}{5 x^{3} - 5 x^{2}} d x} = \frac{x \left(- \ln{\left(\left|{x}\right| \right)} + \ln{\left(\left|{x - 1}\right| \right)}\right) + 1}{5 x}$$

積分定数を加える:

$$\int{\frac{1}{5 x^{3} - 5 x^{2}} d x} = \frac{x \left(- \ln{\left(\left|{x}\right| \right)} + \ln{\left(\left|{x - 1}\right| \right)}\right) + 1}{5 x}+C$$

$$$\int \frac{1}{5 x^{3} - 5 x^{2}}\, dx = \frac{x \left(- \ln\left(\left|{x}\right|\right) + \ln\left(\left|{x - 1}\right|\right)\right) + 1}{5 x} + C$$$A