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

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

对 $$$c=-1$$$ 和 $$$f{\left(x \right)} = \operatorname{atan}{\left(5 x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

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

设$$$u=5 x$$$。

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

积分变为

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

对 $$$c=\frac{1}{5}$$$ 和 $$$f{\left(u \right)} = \operatorname{atan}{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

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

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 »)。

积分变为

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

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，

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

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

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

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

$$- \frac{u \operatorname{atan}{\left(u \right)}}{5} + \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{10} = - \frac{u \operatorname{atan}{\left(u \right)}}{5} + \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{10}$$

回忆一下 $$$v=u^{2} + 1$$$:

$$- \frac{u \operatorname{atan}{\left(u \right)}}{5} + \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{10} = - \frac{u \operatorname{atan}{\left(u \right)}}{5} + \frac{\ln{\left(\left|{{\color{red}{\left(u^{2} + 1\right)}}}\right| \right)}}{10}$$

回忆一下 $$$u=5 x$$$:

$$\frac{\ln{\left(1 + {\color{red}{u}}^{2} \right)}}{10} - \frac{{\color{red}{u}} \operatorname{atan}{\left({\color{red}{u}} \right)}}{5} = \frac{\ln{\left(1 + {\color{red}{\left(5 x\right)}}^{2} \right)}}{10} - \frac{{\color{red}{\left(5 x\right)}} \operatorname{atan}{\left({\color{red}{\left(5 x\right)}} \right)}}{5}$$

因此，

$$\int{\left(- \operatorname{atan}{\left(5 x \right)}\right)d x} = - x \operatorname{atan}{\left(5 x \right)} + \frac{\ln{\left(25 x^{2} + 1 \right)}}{10}$$

加上积分常数：

$$\int{\left(- \operatorname{atan}{\left(5 x \right)}\right)d x} = - x \operatorname{atan}{\left(5 x \right)} + \frac{\ln{\left(25 x^{2} + 1 \right)}}{10}+C$$

$$$\int \left(- \operatorname{atan}{\left(5 x \right)}\right)\, dx = \left(- x \operatorname{atan}{\left(5 x \right)} + \frac{\ln\left(25 x^{2} + 1\right)}{10}\right) + C$$$A