$$$e^{- x} \operatorname{atan}{\left(e^{x} \right)}$$$ 的積分

此計算器將求出 $$$e^{- x} \operatorname{atan}{\left(e^{x} \right)}$$$ 的不定積分(原函數),並顯示步驟。

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您的輸入

$$$\int e^{- x} \operatorname{atan}{\left(e^{x} \right)}\, dx$$$

解答

$$$u=e^{x}$$$

$$$du=\left(e^{x}\right)^{\prime }dx = e^{x} dx$$$ (步驟見»),並可得 $$$e^{x} dx = du$$$

因此,

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

$$$v=\frac{1}{u}$$$

$$$dv=\left(\frac{1}{u}\right)^{\prime }du = - \frac{1}{u^{2}} du$$$ (步驟見»),並可得 $$$\frac{du}{u^{2}} = - dv$$$

該積分變為

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

套用常數倍法則 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$,使用 $$$c=-1$$$$$$f{\left(v \right)} = \operatorname{acot}{\left(v \right)}$$$

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

對於積分 $$$\int{\operatorname{acot}{\left(v \right)} d v}$$$,使用分部積分法 $$$\int \operatorname{m} \operatorname{dy} = \operatorname{m}\operatorname{y} - \int \operatorname{y} \operatorname{dm}$$$

$$$\operatorname{m}=\operatorname{acot}{\left(v \right)}$$$$$$\operatorname{dy}=dv$$$

$$$\operatorname{dm}=\left(\operatorname{acot}{\left(v \right)}\right)^{\prime }dv=- \frac{1}{v^{2} + 1} dv$$$(步驟見 »),且 $$$\operatorname{y}=\int{1 d v}=v$$$(步驟見 »)。

該積分變為

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

$$$w=v^{2} + 1$$$

$$$dw=\left(v^{2} + 1\right)^{\prime }dv = 2 v dv$$$ (步驟見»),並可得 $$$v dv = \frac{dw}{2}$$$

該積分可改寫為

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

套用常數倍法則 $$$\int c f{\left(w \right)}\, dw = c \int f{\left(w \right)}\, dw$$$,使用 $$$c=- \frac{1}{2}$$$$$$f{\left(w \right)} = \frac{1}{w}$$$

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

$$$\frac{1}{w}$$$ 的積分是 $$$\int{\frac{1}{w} d w} = \ln{\left(\left|{w}\right| \right)}$$$

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

回顧一下 $$$w=v^{2} + 1$$$

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

回顧一下 $$$v=\frac{1}{u}$$$

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

回顧一下 $$$u=e^{x}$$$

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

因此,

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

化簡:

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

加上積分常數:

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

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


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