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