$$$e^{- x} \operatorname{atan}{\left(e^{x} \right)}$$$ 的积分
您的输入
求$$$\int e^{- x} \operatorname{atan}{\left(e^{x} \right)}\, dx$$$。
解答
设$$$u=e^{x}$$$。
则$$$du=\left(e^{x}\right)^{\prime }dx = e^{x} dx$$$ (步骤见»),并有$$$e^{x} dx = du$$$。
因此,
$${\color{red}{\int{e^{- x} \operatorname{atan}{\left(e^{x} \right)} d x}}} = {\color{red}{\int{\frac{\operatorname{atan}{\left(u \right)}}{u^{2}} d u}}}$$
设$$$v=\frac{1}{u}$$$。
则$$$dv=\left(\frac{1}{u}\right)^{\prime }du = - \frac{1}{u^{2}} du$$$ (步骤见»),并有$$$\frac{du}{u^{2}} = - dv$$$。
积分变为
$${\color{red}{\int{\frac{\operatorname{atan}{\left(u \right)}}{u^{2}} d u}}} = {\color{red}{\int{\left(- \operatorname{acot}{\left(v \right)}\right)d v}}}$$
对 $$$c=-1$$$ 和 $$$f{\left(v \right)} = \operatorname{acot}{\left(v \right)}$$$ 应用常数倍法则 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$:
$${\color{red}{\int{\left(- \operatorname{acot}{\left(v \right)}\right)d v}}} = {\color{red}{\left(- \int{\operatorname{acot}{\left(v \right)} d v}\right)}}$$
对于积分$$$\int{\operatorname{acot}{\left(v \right)} d v}$$$,使用分部积分法$$$\int \operatorname{m} \operatorname{dy} = \operatorname{m}\operatorname{y} - \int \operatorname{y} \operatorname{dm}$$$。
设 $$$\operatorname{m}=\operatorname{acot}{\left(v \right)}$$$ 和 $$$\operatorname{dy}=dv$$$。
则 $$$\operatorname{dm}=\left(\operatorname{acot}{\left(v \right)}\right)^{\prime }dv=- \frac{1}{v^{2} + 1} dv$$$ (步骤见 »),并且 $$$\operatorname{y}=\int{1 d v}=v$$$ (步骤见 »)。
因此,
$$- {\color{red}{\int{\operatorname{acot}{\left(v \right)} d v}}}=- {\color{red}{\left(\operatorname{acot}{\left(v \right)} \cdot v-\int{v \cdot \left(- \frac{1}{v^{2} + 1}\right) d v}\right)}}=- {\color{red}{\left(v \operatorname{acot}{\left(v \right)} - \int{\left(- \frac{v}{v^{2} + 1}\right)d v}\right)}}$$
设$$$w=v^{2} + 1$$$。
则$$$dw=\left(v^{2} + 1\right)^{\prime }dv = 2 v dv$$$ (步骤见»),并有$$$v dv = \frac{dw}{2}$$$。
因此,
$$- v \operatorname{acot}{\left(v \right)} + {\color{red}{\int{\left(- \frac{v}{v^{2} + 1}\right)d v}}} = - v \operatorname{acot}{\left(v \right)} + {\color{red}{\int{\left(- \frac{1}{2 w}\right)d w}}}$$
对 $$$c=- \frac{1}{2}$$$ 和 $$$f{\left(w \right)} = \frac{1}{w}$$$ 应用常数倍法则 $$$\int c f{\left(w \right)}\, dw = c \int f{\left(w \right)}\, dw$$$:
$$- v \operatorname{acot}{\left(v \right)} + {\color{red}{\int{\left(- \frac{1}{2 w}\right)d w}}} = - v \operatorname{acot}{\left(v \right)} + {\color{red}{\left(- \frac{\int{\frac{1}{w} d w}}{2}\right)}}$$
$$$\frac{1}{w}$$$ 的积分为 $$$\int{\frac{1}{w} d w} = \ln{\left(\left|{w}\right| \right)}$$$:
$$- v \operatorname{acot}{\left(v \right)} - \frac{{\color{red}{\int{\frac{1}{w} d w}}}}{2} = - v \operatorname{acot}{\left(v \right)} - \frac{{\color{red}{\ln{\left(\left|{w}\right| \right)}}}}{2}$$
回忆一下 $$$w=v^{2} + 1$$$:
$$- v \operatorname{acot}{\left(v \right)} - \frac{\ln{\left(\left|{{\color{red}{w}}}\right| \right)}}{2} = - v \operatorname{acot}{\left(v \right)} - \frac{\ln{\left(\left|{{\color{red}{\left(v^{2} + 1\right)}}}\right| \right)}}{2}$$
回忆一下 $$$v=\frac{1}{u}$$$:
$$- \frac{\ln{\left(1 + {\color{red}{v}}^{2} \right)}}{2} - {\color{red}{v}} \operatorname{acot}{\left({\color{red}{v}} \right)} = - \frac{\ln{\left(1 + {\color{red}{\frac{1}{u}}}^{2} \right)}}{2} - {\color{red}{\frac{1}{u}}} \operatorname{acot}{\left({\color{red}{\frac{1}{u}}} \right)}$$
回忆一下 $$$u=e^{x}$$$:
$$- \frac{\ln{\left(1 + {\color{red}{u}}^{-2} \right)}}{2} - {\color{red}{u}}^{-1} \operatorname{acot}{\left({\color{red}{u}}^{-1} \right)} = - \frac{\ln{\left(1 + {\color{red}{e^{x}}}^{-2} \right)}}{2} - {\color{red}{e^{x}}}^{-1} \operatorname{acot}{\left({\color{red}{e^{x}}}^{-1} \right)}$$
因此,
$$\int{e^{- x} \operatorname{atan}{\left(e^{x} \right)} d x} = - \frac{\ln{\left(1 + e^{- 2 x} \right)}}{2} - e^{- x} \operatorname{acot}{\left(e^{- x} \right)}$$
化简:
$$\int{e^{- x} \operatorname{atan}{\left(e^{x} \right)} d x} = x - \frac{\ln{\left(e^{2 x} + 1 \right)}}{2} - e^{- x} \operatorname{atan}{\left(e^{x} \right)}$$
加上积分常数:
$$\int{e^{- x} \operatorname{atan}{\left(e^{x} \right)} d x} = x - \frac{\ln{\left(e^{2 x} + 1 \right)}}{2} - e^{- x} \operatorname{atan}{\left(e^{x} \right)}+C$$
答案
$$$\int e^{- x} \operatorname{atan}{\left(e^{x} \right)}\, dx = \left(x - \frac{\ln\left(e^{2 x} + 1\right)}{2} - e^{- x} \operatorname{atan}{\left(e^{x} \right)}\right) + C$$$A