$$$w^{2} \ln\left(w\right)$$$ 的积分
您的输入
求$$$\int w^{2} \ln\left(w\right)\, dw$$$。
解答
对于积分$$$\int{w^{2} \ln{\left(w \right)} d w}$$$,使用分部积分法$$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。
设 $$$\operatorname{u}=\ln{\left(w \right)}$$$ 和 $$$\operatorname{dv}=w^{2} dw$$$。
则 $$$\operatorname{du}=\left(\ln{\left(w \right)}\right)^{\prime }dw=\frac{dw}{w}$$$ (步骤见 »),并且 $$$\operatorname{v}=\int{w^{2} d w}=\frac{w^{3}}{3}$$$ (步骤见 »)。
积分变为
$${\color{red}{\int{w^{2} \ln{\left(w \right)} d w}}}={\color{red}{\left(\ln{\left(w \right)} \cdot \frac{w^{3}}{3}-\int{\frac{w^{3}}{3} \cdot \frac{1}{w} d w}\right)}}={\color{red}{\left(\frac{w^{3} \ln{\left(w \right)}}{3} - \int{\frac{w^{2}}{3} d w}\right)}}$$
对 $$$c=\frac{1}{3}$$$ 和 $$$f{\left(w \right)} = w^{2}$$$ 应用常数倍法则 $$$\int c f{\left(w \right)}\, dw = c \int f{\left(w \right)}\, dw$$$:
$$\frac{w^{3} \ln{\left(w \right)}}{3} - {\color{red}{\int{\frac{w^{2}}{3} d w}}} = \frac{w^{3} \ln{\left(w \right)}}{3} - {\color{red}{\left(\frac{\int{w^{2} d w}}{3}\right)}}$$
应用幂法则 $$$\int w^{n}\, dw = \frac{w^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,其中 $$$n=2$$$:
$$\frac{w^{3} \ln{\left(w \right)}}{3} - \frac{{\color{red}{\int{w^{2} d w}}}}{3}=\frac{w^{3} \ln{\left(w \right)}}{3} - \frac{{\color{red}{\frac{w^{1 + 2}}{1 + 2}}}}{3}=\frac{w^{3} \ln{\left(w \right)}}{3} - \frac{{\color{red}{\left(\frac{w^{3}}{3}\right)}}}{3}$$
因此,
$$\int{w^{2} \ln{\left(w \right)} d w} = \frac{w^{3} \ln{\left(w \right)}}{3} - \frac{w^{3}}{9}$$
化简:
$$\int{w^{2} \ln{\left(w \right)} d w} = \frac{w^{3} \left(3 \ln{\left(w \right)} - 1\right)}{9}$$
加上积分常数:
$$\int{w^{2} \ln{\left(w \right)} d w} = \frac{w^{3} \left(3 \ln{\left(w \right)} - 1\right)}{9}+C$$
答案
$$$\int w^{2} \ln\left(w\right)\, dw = \frac{w^{3} \left(3 \ln\left(w\right) - 1\right)}{9} + C$$$A