$$$x \operatorname{atan}{\left(x \right)}$$$ 的积分

求$$$\int x \operatorname{atan}{\left(x \right)}\, dx$$$。

对于积分$$$\int{x \operatorname{atan}{\left(x \right)} d x}$$$，使用分部积分法$$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。

设 $$$\operatorname{u}=\operatorname{atan}{\left(x \right)}$$$ 和 $$$\operatorname{dv}=x dx$$$。

则 $$$\operatorname{du}=\left(\operatorname{atan}{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x^{2} + 1}$$$ (步骤见 »)，并且 $$$\operatorname{v}=\int{x d x}=\frac{x^{2}}{2}$$$ (步骤见 »)。

因此，

$${\color{red}{\int{x \operatorname{atan}{\left(x \right)} d x}}}={\color{red}{\left(\operatorname{atan}{\left(x \right)} \cdot \frac{x^{2}}{2}-\int{\frac{x^{2}}{2} \cdot \frac{1}{x^{2} + 1} d x}\right)}}={\color{red}{\left(\frac{x^{2} \operatorname{atan}{\left(x \right)}}{2} - \int{\frac{x^{2}}{2 x^{2} + 2} d x}\right)}}$$

化简被积函数:

$$\frac{x^{2} \operatorname{atan}{\left(x \right)}}{2} - {\color{red}{\int{\frac{x^{2}}{2 x^{2} + 2} d x}}} = \frac{x^{2} \operatorname{atan}{\left(x \right)}}{2} - {\color{red}{\int{\frac{x^{2}}{2 \left(x^{2} + 1\right)} d x}}}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(x \right)} = \frac{x^{2}}{x^{2} + 1}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$\frac{x^{2} \operatorname{atan}{\left(x \right)}}{2} - {\color{red}{\int{\frac{x^{2}}{2 \left(x^{2} + 1\right)} d x}}} = \frac{x^{2} \operatorname{atan}{\left(x \right)}}{2} - {\color{red}{\left(\frac{\int{\frac{x^{2}}{x^{2} + 1} d x}}{2}\right)}}$$

改写并拆分该分式:

$$\frac{x^{2} \operatorname{atan}{\left(x \right)}}{2} - \frac{{\color{red}{\int{\frac{x^{2}}{x^{2} + 1} d x}}}}{2} = \frac{x^{2} \operatorname{atan}{\left(x \right)}}{2} - \frac{{\color{red}{\int{\left(1 - \frac{1}{x^{2} + 1}\right)d x}}}}{2}$$

逐项积分：

$$\frac{x^{2} \operatorname{atan}{\left(x \right)}}{2} - \frac{{\color{red}{\int{\left(1 - \frac{1}{x^{2} + 1}\right)d x}}}}{2} = \frac{x^{2} \operatorname{atan}{\left(x \right)}}{2} - \frac{{\color{red}{\left(\int{1 d x} - \int{\frac{1}{x^{2} + 1} d x}\right)}}}{2}$$

应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=1$$$：

$$\frac{x^{2} \operatorname{atan}{\left(x \right)}}{2} + \frac{\int{\frac{1}{x^{2} + 1} d x}}{2} - \frac{{\color{red}{\int{1 d x}}}}{2} = \frac{x^{2} \operatorname{atan}{\left(x \right)}}{2} + \frac{\int{\frac{1}{x^{2} + 1} d x}}{2} - \frac{{\color{red}{x}}}{2}$$

$$$\frac{1}{x^{2} + 1}$$$ 的积分为 $$$\int{\frac{1}{x^{2} + 1} d x} = \operatorname{atan}{\left(x \right)}$$$:

$$\frac{x^{2} \operatorname{atan}{\left(x \right)}}{2} - \frac{x}{2} + \frac{{\color{red}{\int{\frac{1}{x^{2} + 1} d x}}}}{2} = \frac{x^{2} \operatorname{atan}{\left(x \right)}}{2} - \frac{x}{2} + \frac{{\color{red}{\operatorname{atan}{\left(x \right)}}}}{2}$$

因此，

$$\int{x \operatorname{atan}{\left(x \right)} d x} = \frac{x^{2} \operatorname{atan}{\left(x \right)}}{2} - \frac{x}{2} + \frac{\operatorname{atan}{\left(x \right)}}{2}$$

化简：

$$\int{x \operatorname{atan}{\left(x \right)} d x} = \frac{x^{2} \operatorname{atan}{\left(x \right)} - x + \operatorname{atan}{\left(x \right)}}{2}$$

加上积分常数：

$$\int{x \operatorname{atan}{\left(x \right)} d x} = \frac{x^{2} \operatorname{atan}{\left(x \right)} - x + \operatorname{atan}{\left(x \right)}}{2}+C$$

$$$\int x \operatorname{atan}{\left(x \right)}\, dx = \frac{x^{2} \operatorname{atan}{\left(x \right)} - x + \operatorname{atan}{\left(x \right)}}{2} + C$$$A