Integral von $$$\sqrt[4]{2} x \sqrt[4]{x^{5}}$$$

Bestimme $$$\int \sqrt[4]{2} x \sqrt[4]{x^{5}}\, dx$$$.

Die Eingabe wird umgeschrieben: $$$\int{\sqrt[4]{2} x \sqrt[4]{x^{5}} d x}=\int{\sqrt[4]{2} x^{\frac{9}{4}} d x}$$$.

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\sqrt[4]{2}$$$ und $$$f{\left(x \right)} = x^{\frac{9}{4}}$$$ an:

$${\color{red}{\int{\sqrt[4]{2} x^{\frac{9}{4}} d x}}} = {\color{red}{\sqrt[4]{2} \int{x^{\frac{9}{4}} d x}}}$$

Wenden Sie die Potenzregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=\frac{9}{4}$$$ an:

$$\sqrt[4]{2} {\color{red}{\int{x^{\frac{9}{4}} d x}}}=\sqrt[4]{2} {\color{red}{\frac{x^{1 + \frac{9}{4}}}{1 + \frac{9}{4}}}}=\sqrt[4]{2} {\color{red}{\left(\frac{4 x^{\frac{13}{4}}}{13}\right)}}$$

Daher,

$$\int{\sqrt[4]{2} x^{\frac{9}{4}} d x} = \frac{4 \sqrt[4]{2} x^{\frac{13}{4}}}{13}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\sqrt[4]{2} x^{\frac{9}{4}} d x} = \frac{4 \sqrt[4]{2} x^{\frac{13}{4}}}{13}+C$$

$$$\int \sqrt[4]{2} x \sqrt[4]{x^{5}}\, dx = \frac{4 \sqrt[4]{2} x^{\frac{13}{4}}}{13} + C$$$A