$$$\operatorname{atan}{\left(\frac{x}{5} \right)}$$$ 的积分

求$$$\int \operatorname{atan}{\left(\frac{x}{5} \right)}\, dx$$$。

设$$$u=\frac{x}{5}$$$。

则$$$du=\left(\frac{x}{5}\right)^{\prime }dx = \frac{dx}{5}$$$ (步骤见»)，并有$$$dx = 5 du$$$。

所以，

$${\color{red}{\int{\operatorname{atan}{\left(\frac{x}{5} \right)} d x}}} = {\color{red}{\int{5 \operatorname{atan}{\left(u \right)} d u}}}$$

对 $$$c=5$$$ 和 $$$f{\left(u \right)} = \operatorname{atan}{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$${\color{red}{\int{5 \operatorname{atan}{\left(u \right)} d u}}} = {\color{red}{\left(5 \int{\operatorname{atan}{\left(u \right)} d u}\right)}}$$

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

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

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

该积分可以改写为

$$5 {\color{red}{\int{\operatorname{atan}{\left(u \right)} d u}}}=5 {\color{red}{\left(\operatorname{atan}{\left(u \right)} \cdot u-\int{u \cdot \frac{1}{u^{2} + 1} d u}\right)}}=5 {\color{red}{\left(u \operatorname{atan}{\left(u \right)} - \int{\frac{u}{u^{2} + 1} d u}\right)}}$$

设$$$v=u^{2} + 1$$$。

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

因此，

$$5 u \operatorname{atan}{\left(u \right)} - 5 {\color{red}{\int{\frac{u}{u^{2} + 1} d u}}} = 5 u \operatorname{atan}{\left(u \right)} - 5 {\color{red}{\int{\frac{1}{2 v} d v}}}$$

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

$$5 u \operatorname{atan}{\left(u \right)} - 5 {\color{red}{\int{\frac{1}{2 v} d v}}} = 5 u \operatorname{atan}{\left(u \right)} - 5 {\color{red}{\left(\frac{\int{\frac{1}{v} d v}}{2}\right)}}$$

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

$$5 u \operatorname{atan}{\left(u \right)} - \frac{5 {\color{red}{\int{\frac{1}{v} d v}}}}{2} = 5 u \operatorname{atan}{\left(u \right)} - \frac{5 {\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{2}$$

回忆一下 $$$v=u^{2} + 1$$$:

$$5 u \operatorname{atan}{\left(u \right)} - \frac{5 \ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{2} = 5 u \operatorname{atan}{\left(u \right)} - \frac{5 \ln{\left(\left|{{\color{red}{\left(u^{2} + 1\right)}}}\right| \right)}}{2}$$

回忆一下 $$$u=\frac{x}{5}$$$:

$$- \frac{5 \ln{\left(1 + {\color{red}{u}}^{2} \right)}}{2} + 5 {\color{red}{u}} \operatorname{atan}{\left({\color{red}{u}} \right)} = - \frac{5 \ln{\left(1 + {\color{red}{\left(\frac{x}{5}\right)}}^{2} \right)}}{2} + 5 {\color{red}{\left(\frac{x}{5}\right)}} \operatorname{atan}{\left({\color{red}{\left(\frac{x}{5}\right)}} \right)}$$

因此，

$$\int{\operatorname{atan}{\left(\frac{x}{5} \right)} d x} = x \operatorname{atan}{\left(\frac{x}{5} \right)} - \frac{5 \ln{\left(\frac{x^{2}}{25} + 1 \right)}}{2}$$

加上积分常数：

$$\int{\operatorname{atan}{\left(\frac{x}{5} \right)} d x} = x \operatorname{atan}{\left(\frac{x}{5} \right)} - \frac{5 \ln{\left(\frac{x^{2}}{25} + 1 \right)}}{2}+C$$

$$$\int \operatorname{atan}{\left(\frac{x}{5} \right)}\, dx = \left(x \operatorname{atan}{\left(\frac{x}{5} \right)} - \frac{5 \ln\left(\frac{x^{2}}{25} + 1\right)}{2}\right) + C$$$A