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