$$$\frac{2 \operatorname{atan}{\left(x \right)}}{x^{2}}$$$ 的积分

求$$$\int \frac{2 \operatorname{atan}{\left(x \right)}}{x^{2}}\, dx$$$。

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

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

设$$$u=\frac{1}{x}$$$。

则$$$du=\left(\frac{1}{x}\right)^{\prime }dx = - \frac{1}{x^{2}} dx$$$ (步骤见»)，并有$$$\frac{dx}{x^{2}} = - du$$$。

所以，

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

对 $$$c=-1$$$ 和 $$$f{\left(u \right)} = \operatorname{acot}{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$$2 {\color{red}{\int{\left(- \operatorname{acot}{\left(u \right)}\right)d u}}} = 2 {\color{red}{\left(- \int{\operatorname{acot}{\left(u \right)} d u}\right)}}$$

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

设 $$$\operatorname{t}=\operatorname{acot}{\left(u \right)}$$$ 和 $$$\operatorname{dv}=du$$$。

则 $$$\operatorname{dt}=\left(\operatorname{acot}{\left(u \right)}\right)^{\prime }du=- \frac{1}{u^{2} + 1} du$$$ (步骤见 »)，并且 $$$\operatorname{v}=\int{1 d u}=u$$$ (步骤见 »)。

所以，

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

设$$$v=u^{2} + 1$$$。

则$$$dv=\left(u^{2} + 1\right)^{\prime }du = 2 u du$$$ (步骤见»)，并有$$$u du = \frac{dv}{2}$$$。

该积分可以改写为

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

对 $$$c=- \frac{1}{2}$$$ 和 $$$f{\left(v \right)} = \frac{1}{v}$$$ 应用常数倍法则 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$：

$$- 2 u \operatorname{acot}{\left(u \right)} + 2 {\color{red}{\int{\left(- \frac{1}{2 v}\right)d v}}} = - 2 u \operatorname{acot}{\left(u \right)} + 2 {\color{red}{\left(- \frac{\int{\frac{1}{v} d v}}{2}\right)}}$$

$$$\frac{1}{v}$$$ 的积分为 $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

$$- 2 u \operatorname{acot}{\left(u \right)} - {\color{red}{\int{\frac{1}{v} d v}}} = - 2 u \operatorname{acot}{\left(u \right)} - {\color{red}{\ln{\left(\left|{v}\right| \right)}}}$$

回忆一下 $$$v=u^{2} + 1$$$:

$$- 2 u \operatorname{acot}{\left(u \right)} - \ln{\left(\left|{{\color{red}{v}}}\right| \right)} = - 2 u \operatorname{acot}{\left(u \right)} - \ln{\left(\left|{{\color{red}{\left(u^{2} + 1\right)}}}\right| \right)}$$

回忆一下 $$$u=\frac{1}{x}$$$:

$$- \ln{\left(1 + {\color{red}{u}}^{2} \right)} - 2 {\color{red}{u}} \operatorname{acot}{\left({\color{red}{u}} \right)} = - \ln{\left(1 + {\color{red}{\frac{1}{x}}}^{2} \right)} - 2 {\color{red}{\frac{1}{x}}} \operatorname{acot}{\left({\color{red}{\frac{1}{x}}} \right)}$$

因此，

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

化简：

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

加上积分常数：

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

$$$\int \frac{2 \operatorname{atan}{\left(x \right)}}{x^{2}}\, dx = \left(2 \ln\left(x\right) - \ln\left(x^{2} + 1\right) - \frac{2 \operatorname{atan}{\left(x \right)}}{x}\right) + C$$$A