$$$e^{- x} \operatorname{atan}{\left(e^{x} \right)}$$$ 的積分
您的輸入
求$$$\int e^{- x} \operatorname{atan}{\left(e^{x} \right)}\, dx$$$。
解答
令 $$$u=e^{x}$$$。
則 $$$du=\left(e^{x}\right)^{\prime }dx = e^{x} dx$$$ (步驟見»),並可得 $$$e^{x} dx = du$$$。
因此,
$${\color{red}{\int{e^{- x} \operatorname{atan}{\left(e^{x} \right)} d x}}} = {\color{red}{\int{\frac{\operatorname{atan}{\left(u \right)}}{u^{2}} d u}}}$$
令 $$$v=\frac{1}{u}$$$。
則 $$$dv=\left(\frac{1}{u}\right)^{\prime }du = - \frac{1}{u^{2}} du$$$ (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 $$$c=-1$$$ 與 $$$f{\left(v \right)} = \operatorname{acot}{\left(v \right)}$$$:
$${\color{red}{\int{\left(- \operatorname{acot}{\left(v \right)}\right)d v}}} = {\color{red}{\left(- \int{\operatorname{acot}{\left(v \right)} d v}\right)}}$$
對於積分 $$$\int{\operatorname{acot}{\left(v \right)} d v}$$$,使用分部積分法 $$$\int \operatorname{m} \operatorname{dy} = \operatorname{m}\operatorname{y} - \int \operatorname{y} \operatorname{dm}$$$。
令 $$$\operatorname{m}=\operatorname{acot}{\left(v \right)}$$$ 與 $$$\operatorname{dy}=dv$$$。
則 $$$\operatorname{dm}=\left(\operatorname{acot}{\left(v \right)}\right)^{\prime }dv=- \frac{1}{v^{2} + 1} dv$$$(步驟見 »),且 $$$\operatorname{y}=\int{1 d v}=v$$$(步驟見 »)。
該積分變為
$$- {\color{red}{\int{\operatorname{acot}{\left(v \right)} d v}}}=- {\color{red}{\left(\operatorname{acot}{\left(v \right)} \cdot v-\int{v \cdot \left(- \frac{1}{v^{2} + 1}\right) d v}\right)}}=- {\color{red}{\left(v \operatorname{acot}{\left(v \right)} - \int{\left(- \frac{v}{v^{2} + 1}\right)d v}\right)}}$$
令 $$$w=v^{2} + 1$$$。
則 $$$dw=\left(v^{2} + 1\right)^{\prime }dv = 2 v dv$$$ (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 $$$c=- \frac{1}{2}$$$ 與 $$$f{\left(w \right)} = \frac{1}{w}$$$:
$$- v \operatorname{acot}{\left(v \right)} + {\color{red}{\int{\left(- \frac{1}{2 w}\right)d w}}} = - v \operatorname{acot}{\left(v \right)} + {\color{red}{\left(- \frac{\int{\frac{1}{w} d w}}{2}\right)}}$$
$$$\frac{1}{w}$$$ 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