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