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