$$$x \ln\left(x^{2}\right)$$$ 的积分
您的输入
求$$$\int x \ln\left(x^{2}\right)\, dx$$$。
解答
输入已重写为:$$$\int{x \ln{\left(x^{2} \right)} d x}=\int{2 x \ln{\left(x \right)} d x}$$$。
对 $$$c=2$$$ 和 $$$f{\left(x \right)} = x \ln{\left(x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$:
$${\color{red}{\int{2 x \ln{\left(x \right)} d x}}} = {\color{red}{\left(2 \int{x \ln{\left(x \right)} d x}\right)}}$$
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 »)。
该积分可以改写为
$$2 {\color{red}{\int{x \ln{\left(x \right)} d x}}}=2 {\color{red}{\left(\ln{\left(x \right)} \cdot \frac{x^{2}}{2}-\int{\frac{x^{2}}{2} \cdot \frac{1}{x} d x}\right)}}=2 {\color{red}{\left(\frac{x^{2} \ln{\left(x \right)}}{2} - \int{\frac{x}{2} d x}\right)}}$$
对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(x \right)} = x$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$:
$$x^{2} \ln{\left(x \right)} - 2 {\color{red}{\int{\frac{x}{2} d x}}} = x^{2} \ln{\left(x \right)} - 2 {\color{red}{\left(\frac{\int{x d x}}{2}\right)}}$$
应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,其中 $$$n=1$$$:
$$x^{2} \ln{\left(x \right)} - {\color{red}{\int{x d x}}}=x^{2} \ln{\left(x \right)} - {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=x^{2} \ln{\left(x \right)} - {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$
因此,
$$\int{2 x \ln{\left(x \right)} d x} = x^{2} \ln{\left(x \right)} - \frac{x^{2}}{2}$$
化简:
$$\int{2 x \ln{\left(x \right)} d x} = x^{2} \left(\ln{\left(x \right)} - \frac{1}{2}\right)$$
加上积分常数:
$$\int{2 x \ln{\left(x \right)} d x} = x^{2} \left(\ln{\left(x \right)} - \frac{1}{2}\right)+C$$
答案
$$$\int x \ln\left(x^{2}\right)\, dx = x^{2} \left(\ln\left(x\right) - \frac{1}{2}\right) + C$$$A