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