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$$$- \frac{19}{x^{2}} - \frac{4}{x^{5}}$$$ 的積分
您的輸入
求$$$\int \left(- \frac{19}{x^{2}} - \frac{4}{x^{5}}\right)\, dx$$$。
解答
逐項積分：
$${\color{red}{\int{\left(- \frac{19}{x^{2}} - \frac{4}{x^{5}}\right)d x}}} = {\color{red}{\left(- \int{\frac{4}{x^{5}} d x} - \int{\frac{19}{x^{2}} d x}\right)}}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=19$$$ 與 $$$f{\left(x \right)} = \frac{1}{x^{2}}$$$：
$$- \int{\frac{4}{x^{5}} d x} - {\color{red}{\int{\frac{19}{x^{2}} d x}}} = - \int{\frac{4}{x^{5}} d x} - {\color{red}{\left(19 \int{\frac{1}{x^{2}} d x}\right)}}$$
套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=-2$$$：
$$- \int{\frac{4}{x^{5}} d x} - 19 {\color{red}{\int{\frac{1}{x^{2}} d x}}}=- \int{\frac{4}{x^{5}} d x} - 19 {\color{red}{\int{x^{-2} d x}}}=- \int{\frac{4}{x^{5}} d x} - 19 {\color{red}{\frac{x^{-2 + 1}}{-2 + 1}}}=- \int{\frac{4}{x^{5}} d x} - 19 {\color{red}{\left(- x^{-1}\right)}}=- \int{\frac{4}{x^{5}} d x} - 19 {\color{red}{\left(- \frac{1}{x}\right)}}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=4$$$ 與 $$$f{\left(x \right)} = \frac{1}{x^{5}}$$$：
$$- {\color{red}{\int{\frac{4}{x^{5}} d x}}} + \frac{19}{x} = - {\color{red}{\left(4 \int{\frac{1}{x^{5}} d x}\right)}} + \frac{19}{x}$$
套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=-5$$$：
$$- 4 {\color{red}{\int{\frac{1}{x^{5}} d x}}} + \frac{19}{x}=- 4 {\color{red}{\int{x^{-5} d x}}} + \frac{19}{x}=- 4 {\color{red}{\frac{x^{-5 + 1}}{-5 + 1}}} + \frac{19}{x}=- 4 {\color{red}{\left(- \frac{x^{-4}}{4}\right)}} + \frac{19}{x}=- 4 {\color{red}{\left(- \frac{1}{4 x^{4}}\right)}} + \frac{19}{x}$$
因此，
$$\int{\left(- \frac{19}{x^{2}} - \frac{4}{x^{5}}\right)d x} = \frac{19}{x} + \frac{1}{x^{4}}$$
加上積分常數：
$$\int{\left(- \frac{19}{x^{2}} - \frac{4}{x^{5}}\right)d x} = \frac{19}{x} + \frac{1}{x^{4}}+C$$
答案
$$$\int \left(- \frac{19}{x^{2}} - \frac{4}{x^{5}}\right)\, dx = \left(\frac{19}{x} + \frac{1}{x^{4}}\right) + C$$$A