Integral von $$$e^{- x} \operatorname{atan}{\left(e^{x} \right)}$$$

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

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

Dann $$$du=\left(e^{x}\right)^{\prime }dx = e^{x} dx$$$ (die Schritte sind » zu sehen), und es gilt $$$e^{x} dx = du$$$.

Das Integral wird zu

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

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

Dann $$$dv=\left(\frac{1}{u}\right)^{\prime }du = - \frac{1}{u^{2}} du$$$ (die Schritte sind » zu sehen), und es gilt $$$\frac{du}{u^{2}} = - dv$$$.

Daher,

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

Wende die Konstantenfaktorregel $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ mit $$$c=-1$$$ und $$$f{\left(v \right)} = \operatorname{acot}{\left(v \right)}$$$ an:

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

Für das Integral $$$\int{\operatorname{acot}{\left(v \right)} d v}$$$ verwenden Sie die partielle Integration $$$\int \operatorname{m} \operatorname{dy} = \operatorname{m}\operatorname{y} - \int \operatorname{y} \operatorname{dm}$$$.

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

Dann gilt $$$\operatorname{dm}=\left(\operatorname{acot}{\left(v \right)}\right)^{\prime }dv=- \frac{1}{v^{2} + 1} dv$$$ (Rechenschritte siehe ») und $$$\operatorname{y}=\int{1 d v}=v$$$ (Rechenschritte siehe »).

Daher,

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

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

Dann $$$dw=\left(v^{2} + 1\right)^{\prime }dv = 2 v dv$$$ (die Schritte sind » zu sehen), und es gilt $$$v dv = \frac{dw}{2}$$$.

Somit,

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

Wende die Konstantenfaktorregel $$$\int c f{\left(w \right)}\, dw = c \int f{\left(w \right)}\, dw$$$ mit $$$c=- \frac{1}{2}$$$ und $$$f{\left(w \right)} = \frac{1}{w}$$$ an:

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

Das Integral von $$$\frac{1}{w}$$$ ist $$$\int{\frac{1}{w} d w} = \ln{\left(\left|{w}\right| \right)}$$$:

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

Zur Erinnerung: $$$w=v^{2} + 1$$$:

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

Zur Erinnerung: $$$v=\frac{1}{u}$$$:

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

Zur Erinnerung: $$$u=e^{x}$$$:

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

Daher,

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

Vereinfachen:

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

Fügen Sie die Integrationskonstante hinzu:

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

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