Integral von $$$\frac{3 e^{\frac{1}{x^{3}}}}{x^{4}}$$$

Bestimme $$$\int \frac{3 e^{\frac{1}{x^{3}}}}{x^{4}}\, dx$$$.

Sei $$$u=x^{3}$$$.

Dann $$$du=\left(x^{3}\right)^{\prime }dx = 3 x^{2} dx$$$ (die Schritte sind » zu sehen), und es gilt $$$x^{2} dx = \frac{du}{3}$$$.

Das Integral wird zu

$${\color{red}{\int{\frac{3 e^{\frac{1}{x^{3}}}}{x^{4}} d x}}} = {\color{red}{\int{\frac{e^{\frac{1}{u}}}{u^{2}} d u}}}$$

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 $$$\frac{du}{u^{2}} = - dv$$$.

Also,

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

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

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

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

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

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

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

Zur Erinnerung: $$$u=x^{3}$$$:

$$- e^{{\color{red}{u}}^{-1}} = - e^{{\color{red}{x^{3}}}^{-1}}$$

Daher,

$$\int{\frac{3 e^{\frac{1}{x^{3}}}}{x^{4}} d x} = - e^{\frac{1}{x^{3}}}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{3 e^{\frac{1}{x^{3}}}}{x^{4}} d x} = - e^{\frac{1}{x^{3}}}+C$$

$$$\int \frac{3 e^{\frac{1}{x^{3}}}}{x^{4}}\, dx = - e^{\frac{1}{x^{3}}} + C$$$A