Integral von $$$\left(4 x - 2\right) e^{x^{2} - x}$$$

Bestimme $$$\int \left(4 x - 2\right) e^{x^{2} - x}\, dx$$$.

Die Eingabe wird umgeschrieben: $$$\int{\left(4 x - 2\right) e^{x^{2} - x} d x}=\int{\left(4 x - 2\right) e^{x \left(x - 1\right)} d x}$$$.

Sei $$$u=x \left(x - 1\right)$$$.

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

Daher,

$${\color{red}{\int{\left(4 x - 2\right) e^{x \left(x - 1\right)} d x}}} = {\color{red}{\int{2 e^{u} d u}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=2$$$ und $$$f{\left(u \right)} = e^{u}$$$ an:

$${\color{red}{\int{2 e^{u} d u}}} = {\color{red}{\left(2 \int{e^{u} d u}\right)}}$$

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

$$2 {\color{red}{\int{e^{u} d u}}} = 2 {\color{red}{e^{u}}}$$

Zur Erinnerung: $$$u=x \left(x - 1\right)$$$:

$$2 e^{{\color{red}{u}}} = 2 e^{{\color{red}{x \left(x - 1\right)}}}$$

Daher,

$$\int{\left(4 x - 2\right) e^{x \left(x - 1\right)} d x} = 2 e^{x \left(x - 1\right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\left(4 x - 2\right) e^{x \left(x - 1\right)} d x} = 2 e^{x \left(x - 1\right)}+C$$

$$$\int \left(4 x - 2\right) e^{x^{2} - x}\, dx = 2 e^{x \left(x - 1\right)} + C$$$A