Integral von $$$\frac{e^{\frac{1}{x}}}{x^{3}}$$$

Bestimme $$$\int \frac{e^{\frac{1}{x}}}{x^{3}}\, dx$$$.

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

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

Also,

$${\color{red}{\int{\frac{e^{\frac{1}{x}}}{x^{3}} d x}}} = {\color{red}{\int{\left(- u e^{u}\right)d u}}}$$

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

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

Für das Integral $$$\int{u e^{u} d u}$$$ verwenden Sie die partielle Integration $$$\int \operatorname{p} \operatorname{dv} = \operatorname{p}\operatorname{v} - \int \operatorname{v} \operatorname{dp}$$$.

Seien $$$\operatorname{p}=u$$$ und $$$\operatorname{dv}=e^{u} du$$$.

Dann gilt $$$\operatorname{dp}=\left(u\right)^{\prime }du=1 du$$$ (Rechenschritte siehe ») und $$$\operatorname{v}=\int{e^{u} d u}=e^{u}$$$ (Rechenschritte siehe »).

Daher,

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

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

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

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

$$e^{{\color{red}{u}}} - {\color{red}{u}} e^{{\color{red}{u}}} = e^{{\color{red}{\frac{1}{x}}}} - {\color{red}{\frac{1}{x}}} e^{{\color{red}{\frac{1}{x}}}}$$

Daher,

$$\int{\frac{e^{\frac{1}{x}}}{x^{3}} d x} = e^{\frac{1}{x}} - \frac{e^{\frac{1}{x}}}{x}$$

Vereinfachen:

$$\int{\frac{e^{\frac{1}{x}}}{x^{3}} d x} = \frac{\left(x - 1\right) e^{\frac{1}{x}}}{x}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{e^{\frac{1}{x}}}{x^{3}} d x} = \frac{\left(x - 1\right) e^{\frac{1}{x}}}{x}+C$$

$$$\int \frac{e^{\frac{1}{x}}}{x^{3}}\, dx = \frac{\left(x - 1\right) e^{\frac{1}{x}}}{x} + C$$$A