$$$\frac{8 \left(x - 1\right)}{x^{2}}$$$ 的積分
您的輸入
求$$$\int \frac{8 \left(x - 1\right)}{x^{2}}\, dx$$$。
解答
已將輸入重寫為：$$$\int{\frac{8 \left(x - 1\right)}{x^{2}} d x}=\int{\frac{8 x - 8}{x^{2}} d x}$$$。
Expand the expression:
$${\color{red}{\int{\frac{8 x - 8}{x^{2}} d x}}} = {\color{red}{\int{\left(\frac{8}{x} - \frac{8}{x^{2}}\right)d x}}}$$
逐項積分：
$${\color{red}{\int{\left(\frac{8}{x} - \frac{8}{x^{2}}\right)d x}}} = {\color{red}{\left(- \int{\frac{8}{x^{2}} d x} + \int{\frac{8}{x} d x}\right)}}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=8$$$ 與 $$$f{\left(x \right)} = \frac{1}{x^{2}}$$$：
$$\int{\frac{8}{x} d x} - {\color{red}{\int{\frac{8}{x^{2}} d x}}} = \int{\frac{8}{x} d x} - {\color{red}{\left(8 \int{\frac{1}{x^{2}} d x}\right)}}$$
套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=-2$$$：
$$\int{\frac{8}{x} d x} - 8 {\color{red}{\int{\frac{1}{x^{2}} d x}}}=\int{\frac{8}{x} d x} - 8 {\color{red}{\int{x^{-2} d x}}}=\int{\frac{8}{x} d x} - 8 {\color{red}{\frac{x^{-2 + 1}}{-2 + 1}}}=\int{\frac{8}{x} d x} - 8 {\color{red}{\left(- x^{-1}\right)}}=\int{\frac{8}{x} d x} - 8 {\color{red}{\left(- \frac{1}{x}\right)}}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=8$$$ 與 $$$f{\left(x \right)} = \frac{1}{x}$$$：
$${\color{red}{\int{\frac{8}{x} d x}}} + \frac{8}{x} = {\color{red}{\left(8 \int{\frac{1}{x} d x}\right)}} + \frac{8}{x}$$
$$$\frac{1}{x}$$$ 的積分是 $$$\int{\frac{1}{x} d x} = \ln{\left(\left|{x}\right| \right)}$$$：
$$8 {\color{red}{\int{\frac{1}{x} d x}}} + \frac{8}{x} = 8 {\color{red}{\ln{\left(\left|{x}\right| \right)}}} + \frac{8}{x}$$
因此，
$$\int{\frac{8 x - 8}{x^{2}} d x} = 8 \ln{\left(\left|{x}\right| \right)} + \frac{8}{x}$$
加上積分常數：
$$\int{\frac{8 x - 8}{x^{2}} d x} = 8 \ln{\left(\left|{x}\right| \right)} + \frac{8}{x}+C$$
答案
$$$\int \frac{8 \left(x - 1\right)}{x^{2}}\, dx = \left(8 \ln\left(\left|{x}\right|\right) + \frac{8}{x}\right) + C$$$A