$$$\frac{x - 1}{x^{2}}$$$の積分

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

Expand the expression:

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

項別に積分せよ:

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

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

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

$$$n=-2$$$ 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:

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

したがって、

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

積分定数を加える:

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

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