$$$x^{2} - \frac{12}{x^{31}}$$$ 的積分
您的輸入
求$$$\int \left(x^{2} - \frac{12}{x^{31}}\right)\, dx$$$。
解答
逐項積分：
$${\color{red}{\int{\left(x^{2} - \frac{12}{x^{31}}\right)d x}}} = {\color{red}{\left(- \int{\frac{12}{x^{31}} d x} + \int{x^{2} d x}\right)}}$$
套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=2$$$：
$$- \int{\frac{12}{x^{31}} d x} + {\color{red}{\int{x^{2} d x}}}=- \int{\frac{12}{x^{31}} d x} + {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=- \int{\frac{12}{x^{31}} d x} + {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=12$$$ 與 $$$f{\left(x \right)} = \frac{1}{x^{31}}$$$：
$$\frac{x^{3}}{3} - {\color{red}{\int{\frac{12}{x^{31}} d x}}} = \frac{x^{3}}{3} - {\color{red}{\left(12 \int{\frac{1}{x^{31}} d x}\right)}}$$
套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=-31$$$：
$$\frac{x^{3}}{3} - 12 {\color{red}{\int{\frac{1}{x^{31}} d x}}}=\frac{x^{3}}{3} - 12 {\color{red}{\int{x^{-31} d x}}}=\frac{x^{3}}{3} - 12 {\color{red}{\frac{x^{-31 + 1}}{-31 + 1}}}=\frac{x^{3}}{3} - 12 {\color{red}{\left(- \frac{x^{-30}}{30}\right)}}=\frac{x^{3}}{3} - 12 {\color{red}{\left(- \frac{1}{30 x^{30}}\right)}}$$
因此，
$$\int{\left(x^{2} - \frac{12}{x^{31}}\right)d x} = \frac{x^{3}}{3} + \frac{2}{5 x^{30}}$$
化簡：
$$\int{\left(x^{2} - \frac{12}{x^{31}}\right)d x} = \frac{5 x^{33} + 6}{15 x^{30}}$$
加上積分常數：
$$\int{\left(x^{2} - \frac{12}{x^{31}}\right)d x} = \frac{5 x^{33} + 6}{15 x^{30}}+C$$
答案
$$$\int \left(x^{2} - \frac{12}{x^{31}}\right)\, dx = \frac{5 x^{33} + 6}{15 x^{30}} + C$$$A