Integral von $$$\frac{2 e^{- \frac{2}{x}}}{x^{2}}$$$

Bestimme $$$\int \frac{2 e^{- \frac{2}{x}}}{x^{2}}\, dx$$$.

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=2$$$ und $$$f{\left(x \right)} = \frac{e^{- \frac{2}{x}}}{x^{2}}$$$ an:

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

Sei $$$u=- \frac{2}{x}$$$.

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

Das Integral wird zu

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

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{2}$$$ und $$$f{\left(u \right)} = e^{u}$$$ an:

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

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: $$$u=- \frac{2}{x}$$$:

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

Daher,

$$\int{\frac{2 e^{- \frac{2}{x}}}{x^{2}} d x} = e^{- \frac{2}{x}}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{2 e^{- \frac{2}{x}}}{x^{2}} d x} = e^{- \frac{2}{x}}+C$$

$$$\int \frac{2 e^{- \frac{2}{x}}}{x^{2}}\, dx = e^{- \frac{2}{x}} + C$$$A