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$$$-8 + \frac{1}{x^{3}}$$$ 的積分
您的輸入
求$$$\int \left(-8 + \frac{1}{x^{3}}\right)\, dx$$$。
解答
逐項積分：
$${\color{red}{\int{\left(-8 + \frac{1}{x^{3}}\right)d x}}} = {\color{red}{\left(- \int{8 d x} + \int{\frac{1}{x^{3}} d x}\right)}}$$
配合 $$$c=8$$$，應用常數法則 $$$\int c\, dx = c x$$$：
$$\int{\frac{1}{x^{3}} d x} - {\color{red}{\int{8 d x}}} = \int{\frac{1}{x^{3}} d x} - {\color{red}{\left(8 x\right)}}$$
套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=-3$$$：
$$- 8 x + {\color{red}{\int{\frac{1}{x^{3}} d x}}}=- 8 x + {\color{red}{\int{x^{-3} d x}}}=- 8 x + {\color{red}{\frac{x^{-3 + 1}}{-3 + 1}}}=- 8 x + {\color{red}{\left(- \frac{x^{-2}}{2}\right)}}=- 8 x + {\color{red}{\left(- \frac{1}{2 x^{2}}\right)}}$$
因此，
$$\int{\left(-8 + \frac{1}{x^{3}}\right)d x} = - 8 x - \frac{1}{2 x^{2}}$$
加上積分常數：
$$\int{\left(-8 + \frac{1}{x^{3}}\right)d x} = - 8 x - \frac{1}{2 x^{2}}+C$$
答案
$$$\int \left(-8 + \frac{1}{x^{3}}\right)\, dx = \left(- 8 x - \frac{1}{2 x^{2}}\right) + C$$$A