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$$$\frac{1}{256 x^{16}}$$$ 的積分
您的輸入
求$$$\int \frac{1}{256 x^{16}}\, dx$$$。
解答
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{1}{256}$$$ 與 $$$f{\left(x \right)} = \frac{1}{x^{16}}$$$：
$${\color{red}{\int{\frac{1}{256 x^{16}} d x}}} = {\color{red}{\left(\frac{\int{\frac{1}{x^{16}} d x}}{256}\right)}}$$
套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=-16$$$：
$$\frac{{\color{red}{\int{\frac{1}{x^{16}} d x}}}}{256}=\frac{{\color{red}{\int{x^{-16} d x}}}}{256}=\frac{{\color{red}{\frac{x^{-16 + 1}}{-16 + 1}}}}{256}=\frac{{\color{red}{\left(- \frac{x^{-15}}{15}\right)}}}{256}=\frac{{\color{red}{\left(- \frac{1}{15 x^{15}}\right)}}}{256}$$
因此，
$$\int{\frac{1}{256 x^{16}} d x} = - \frac{1}{3840 x^{15}}$$
加上積分常數：
$$\int{\frac{1}{256 x^{16}} d x} = - \frac{1}{3840 x^{15}}+C$$
答案
$$$\int \frac{1}{256 x^{16}}\, dx = - \frac{1}{3840 x^{15}} + C$$$A