$$$\frac{x - 1}{x^{4}}$$$の積分

$$$\int \frac{x - 1}{x^{4}}\, dx$$$ を求めよ。

Expand the expression:

$${\color{red}{\int{\frac{x - 1}{x^{4}} d x}}} = {\color{red}{\int{\left(\frac{1}{x^{3}} - \frac{1}{x^{4}}\right)d x}}}$$

項別に積分せよ:

$${\color{red}{\int{\left(\frac{1}{x^{3}} - \frac{1}{x^{4}}\right)d x}}} = {\color{red}{\left(- \int{\frac{1}{x^{4}} d x} + \int{\frac{1}{x^{3}} d x}\right)}}$$

$$$n=-3$$$ 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:

$$- \int{\frac{1}{x^{4}} d x} + {\color{red}{\int{\frac{1}{x^{3}} d x}}}=- \int{\frac{1}{x^{4}} d x} + {\color{red}{\int{x^{-3} d x}}}=- \int{\frac{1}{x^{4}} d x} + {\color{red}{\frac{x^{-3 + 1}}{-3 + 1}}}=- \int{\frac{1}{x^{4}} d x} + {\color{red}{\left(- \frac{x^{-2}}{2}\right)}}=- \int{\frac{1}{x^{4}} d x} + {\color{red}{\left(- \frac{1}{2 x^{2}}\right)}}$$

$$$n=-4$$$ 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:

$$- {\color{red}{\int{\frac{1}{x^{4}} d x}}} - \frac{1}{2 x^{2}}=- {\color{red}{\int{x^{-4} d x}}} - \frac{1}{2 x^{2}}=- {\color{red}{\frac{x^{-4 + 1}}{-4 + 1}}} - \frac{1}{2 x^{2}}=- {\color{red}{\left(- \frac{x^{-3}}{3}\right)}} - \frac{1}{2 x^{2}}=- {\color{red}{\left(- \frac{1}{3 x^{3}}\right)}} - \frac{1}{2 x^{2}}$$

したがって、

$$\int{\frac{x - 1}{x^{4}} d x} = - \frac{1}{2 x^{2}} + \frac{1}{3 x^{3}}$$

簡単化せよ:

$$\int{\frac{x - 1}{x^{4}} d x} = \frac{2 - 3 x}{6 x^{3}}$$

積分定数を加える:

$$\int{\frac{x - 1}{x^{4}} d x} = \frac{2 - 3 x}{6 x^{3}}+C$$

$$$\int \frac{x - 1}{x^{4}}\, dx = \frac{2 - 3 x}{6 x^{3}} + C$$$A