$$$\ln\left(\frac{x^{n}}{x}\right)$$$ 关于$$$x$$$的积分
您的输入
求$$$\int \ln\left(\frac{x^{n}}{x}\right)\, dx$$$。
解答
输入已重写为：$$$\int{\ln{\left(\frac{x^{n}}{x} \right)} d x}=\int{\left(n - 1\right) \ln{\left(x \right)} d x}$$$。
对 $$$c=n - 1$$$ 和 $$$f{\left(x \right)} = \ln{\left(x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：
$${\color{red}{\int{\left(n - 1\right) \ln{\left(x \right)} d x}}} = {\color{red}{\left(n - 1\right) \int{\ln{\left(x \right)} d x}}}$$
对于积分$$$\int{\ln{\left(x \right)} d x}$$$，使用分部积分法$$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。
设 $$$\operatorname{u}=\ln{\left(x \right)}$$$ 和 $$$\operatorname{dv}=dx$$$。
则 $$$\operatorname{du}=\left(\ln{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x}$$$ (步骤见 »)，并且 $$$\operatorname{v}=\int{1 d x}=x$$$ (步骤见 »)。
因此，
$$\left(n - 1\right) {\color{red}{\int{\ln{\left(x \right)} d x}}}=\left(n - 1\right) {\color{red}{\left(\ln{\left(x \right)} \cdot x-\int{x \cdot \frac{1}{x} d x}\right)}}=\left(n - 1\right) {\color{red}{\left(x \ln{\left(x \right)} - \int{1 d x}\right)}}$$
应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=1$$$：
$$\left(n - 1\right) \left(x \ln{\left(x \right)} - {\color{red}{\int{1 d x}}}\right) = \left(n - 1\right) \left(x \ln{\left(x \right)} - {\color{red}{x}}\right)$$
因此，
$$\int{\left(n - 1\right) \ln{\left(x \right)} d x} = \left(n - 1\right) \left(x \ln{\left(x \right)} - x\right)$$
化简：
$$\int{\left(n - 1\right) \ln{\left(x \right)} d x} = x \left(n - 1\right) \left(\ln{\left(x \right)} - 1\right)$$
加上积分常数：
$$\int{\left(n - 1\right) \ln{\left(x \right)} d x} = x \left(n - 1\right) \left(\ln{\left(x \right)} - 1\right)+C$$
答案
$$$\int \ln\left(\frac{x^{n}}{x}\right)\, dx = x \left(n - 1\right) \left(\ln\left(x\right) - 1\right) + C$$$A