Integral von $$$\frac{e^{\operatorname{atan}{\left(x \right)}}}{x^{2} + 1}$$$

Bestimme $$$\int \frac{e^{\operatorname{atan}{\left(x \right)}}}{x^{2} + 1}\, dx$$$.

Sei $$$u=\operatorname{atan}{\left(x \right)}$$$.

Dann $$$du=\left(\operatorname{atan}{\left(x \right)}\right)^{\prime }dx = \frac{dx}{x^{2} + 1}$$$ (die Schritte sind » zu sehen), und es gilt $$$\frac{dx}{x^{2} + 1} = du$$$.

Das Integral wird zu

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

Das Integral der Exponentialfunktion lautet $$$\int{e^{u} d u} = e^{u}$$$:

$${\color{red}{\int{e^{u} d u}}} = {\color{red}{e^{u}}}$$

Zur Erinnerung: $$$u=\operatorname{atan}{\left(x \right)}$$$:

$$e^{{\color{red}{u}}} = e^{{\color{red}{\operatorname{atan}{\left(x \right)}}}}$$

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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