Integral von $$$\frac{1}{\sqrt[3]{x - 2}}$$$

Bestimme $$$\int \frac{1}{\sqrt[3]{x - 2}}\, dx$$$.

Sei $$$u=x - 2$$$.

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

Also,

$${\color{red}{\int{\frac{1}{\sqrt[3]{x - 2}} d x}}} = {\color{red}{\int{\frac{1}{\sqrt[3]{u}} d u}}}$$

Wenden Sie die Potenzregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=- \frac{1}{3}$$$ an:

$${\color{red}{\int{\frac{1}{\sqrt[3]{u}} d u}}}={\color{red}{\int{u^{- \frac{1}{3}} d u}}}={\color{red}{\frac{u^{- \frac{1}{3} + 1}}{- \frac{1}{3} + 1}}}={\color{red}{\left(\frac{3 u^{\frac{2}{3}}}{2}\right)}}$$

Zur Erinnerung: $$$u=x - 2$$$:

$$\frac{3 {\color{red}{u}}^{\frac{2}{3}}}{2} = \frac{3 {\color{red}{\left(x - 2\right)}}^{\frac{2}{3}}}{2}$$

Daher,

$$\int{\frac{1}{\sqrt[3]{x - 2}} d x} = \frac{3 \left(x - 2\right)^{\frac{2}{3}}}{2}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{1}{\sqrt[3]{x - 2}} d x} = \frac{3 \left(x - 2\right)^{\frac{2}{3}}}{2}+C$$

$$$\int \frac{1}{\sqrt[3]{x - 2}}\, dx = \frac{3 \left(x - 2\right)^{\frac{2}{3}}}{2} + C$$$A