Integralen av $$$\frac{\left(\sqrt[3]{x} - 4\right)^{5}}{6 x^{\frac{2}{3}}}$$$

Bestäm $$$\int \frac{\left(\sqrt[3]{x} - 4\right)^{5}}{6 x^{\frac{2}{3}}}\, dx$$$.

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=\frac{1}{6}$$$ och $$$f{\left(x \right)} = \frac{\left(\sqrt[3]{x} - 4\right)^{5}}{x^{\frac{2}{3}}}$$$:

$${\color{red}{\int{\frac{\left(\sqrt[3]{x} - 4\right)^{5}}{6 x^{\frac{2}{3}}} d x}}} = {\color{red}{\left(\frac{\int{\frac{\left(\sqrt[3]{x} - 4\right)^{5}}{x^{\frac{2}{3}}} d x}}{6}\right)}}$$

Låt $$$u=\sqrt[3]{x} - 4$$$ vara.

Då $$$du=\left(\sqrt[3]{x} - 4\right)^{\prime }dx = \frac{1}{3 x^{\frac{2}{3}}} dx$$$ (stegen kan ses »), och vi har att $$$\frac{dx}{x^{\frac{2}{3}}} = 3 du$$$.

Integralen blir

$$\frac{{\color{red}{\int{\frac{\left(\sqrt[3]{x} - 4\right)^{5}}{x^{\frac{2}{3}}} d x}}}}{6} = \frac{{\color{red}{\int{3 u^{5} d u}}}}{6}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=3$$$ och $$$f{\left(u \right)} = u^{5}$$$:

$$\frac{{\color{red}{\int{3 u^{5} d u}}}}{6} = \frac{{\color{red}{\left(3 \int{u^{5} d u}\right)}}}{6}$$

Tillämpa potensregeln $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=5$$$:

$$\frac{{\color{red}{\int{u^{5} d u}}}}{2}=\frac{{\color{red}{\frac{u^{1 + 5}}{1 + 5}}}}{2}=\frac{{\color{red}{\left(\frac{u^{6}}{6}\right)}}}{2}$$

Kom ihåg att $$$u=\sqrt[3]{x} - 4$$$:

$$\frac{{\color{red}{u}}^{6}}{12} = \frac{{\color{red}{\left(\sqrt[3]{x} - 4\right)}}^{6}}{12}$$

Alltså,

$$\int{\frac{\left(\sqrt[3]{x} - 4\right)^{5}}{6 x^{\frac{2}{3}}} d x} = \frac{\left(\sqrt[3]{x} - 4\right)^{6}}{12}$$

Lägg till integrationskonstanten:

$$\int{\frac{\left(\sqrt[3]{x} - 4\right)^{5}}{6 x^{\frac{2}{3}}} d x} = \frac{\left(\sqrt[3]{x} - 4\right)^{6}}{12}+C$$

$$$\int \frac{\left(\sqrt[3]{x} - 4\right)^{5}}{6 x^{\frac{2}{3}}}\, dx = \frac{\left(\sqrt[3]{x} - 4\right)^{6}}{12} + C$$$A