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$$$x \ln\left(x\right) - x$$$ 的积分
您的输入
求$$$\int \left(x \ln\left(x\right) - x\right)\, dx$$$。
解答
逐项积分：
$${\color{red}{\int{\left(x \ln{\left(x \right)} - x\right)d x}}} = {\color{red}{\left(- \int{x d x} + \int{x \ln{\left(x \right)} d x}\right)}}$$
应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=1$$$：
$$\int{x \ln{\left(x \right)} d x} - {\color{red}{\int{x d x}}}=\int{x \ln{\left(x \right)} d x} - {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=\int{x \ln{\left(x \right)} d x} - {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$
对于积分$$$\int{x \ln{\left(x \right)} d x}$$$，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 $$$\operatorname{dv}=x dx$$$。
则 $$$\operatorname{du}=\left(\ln{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x}$$$ (步骤见 »)，并且 $$$\operatorname{v}=\int{x d x}=\frac{x^{2}}{2}$$$ (步骤见 »)。
所以，
$$- \frac{x^{2}}{2} + {\color{red}{\int{x \ln{\left(x \right)} d x}}}=- \frac{x^{2}}{2} + {\color{red}{\left(\ln{\left(x \right)} \cdot \frac{x^{2}}{2}-\int{\frac{x^{2}}{2} \cdot \frac{1}{x} d x}\right)}}=- \frac{x^{2}}{2} + {\color{red}{\left(\frac{x^{2} \ln{\left(x \right)}}{2} - \int{\frac{x}{2} d x}\right)}}$$
对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(x \right)} = x$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：
$$\frac{x^{2} \ln{\left(x \right)}}{2} - \frac{x^{2}}{2} - {\color{red}{\int{\frac{x}{2} d x}}} = \frac{x^{2} \ln{\left(x \right)}}{2} - \frac{x^{2}}{2} - {\color{red}{\left(\frac{\int{x d x}}{2}\right)}}$$
应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=1$$$：
$$\frac{x^{2} \ln{\left(x \right)}}{2} - \frac{x^{2}}{2} - \frac{{\color{red}{\int{x d x}}}}{2}=\frac{x^{2} \ln{\left(x \right)}}{2} - \frac{x^{2}}{2} - \frac{{\color{red}{\frac{x^{1 + 1}}{1 + 1}}}}{2}=\frac{x^{2} \ln{\left(x \right)}}{2} - \frac{x^{2}}{2} - \frac{{\color{red}{\left(\frac{x^{2}}{2}\right)}}}{2}$$
因此，
$$\int{\left(x \ln{\left(x \right)} - x\right)d x} = \frac{x^{2} \ln{\left(x \right)}}{2} - \frac{3 x^{2}}{4}$$
化简：
$$\int{\left(x \ln{\left(x \right)} - x\right)d x} = \frac{x^{2} \left(2 \ln{\left(x \right)} - 3\right)}{4}$$
加上积分常数：
$$\int{\left(x \ln{\left(x \right)} - x\right)d x} = \frac{x^{2} \left(2 \ln{\left(x \right)} - 3\right)}{4}+C$$
答案
$$$\int \left(x \ln\left(x\right) - x\right)\, dx = \frac{x^{2} \left(2 \ln\left(x\right) - 3\right)}{4} + C$$$A