Integral de $$$5 \sqrt[3]{2 x + 4}$$$

Encontre $$$\int 5 \sqrt[3]{2 x + 4}\, dx$$$.

Simplifique o integrando:

$${\color{red}{\int{5 \sqrt[3]{2 x + 4} d x}}} = {\color{red}{\int{5 \sqrt[3]{2} \sqrt[3]{x + 2} d x}}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=5 \sqrt[3]{2}$$$ e $$$f{\left(x \right)} = \sqrt[3]{x + 2}$$$:

$${\color{red}{\int{5 \sqrt[3]{2} \sqrt[3]{x + 2} d x}}} = {\color{red}{\left(5 \sqrt[3]{2} \int{\sqrt[3]{x + 2} d x}\right)}}$$

Seja $$$u=x + 2$$$.

Então $$$du=\left(x + 2\right)^{\prime }dx = 1 dx$$$ (veja os passos »), e obtemos $$$dx = du$$$.

A integral torna-se

$$5 \sqrt[3]{2} {\color{red}{\int{\sqrt[3]{x + 2} d x}}} = 5 \sqrt[3]{2} {\color{red}{\int{\sqrt[3]{u} d u}}}$$

Aplique a regra da potência $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=\frac{1}{3}$$$:

$$5 \sqrt[3]{2} {\color{red}{\int{\sqrt[3]{u} d u}}}=5 \sqrt[3]{2} {\color{red}{\int{u^{\frac{1}{3}} d u}}}=5 \sqrt[3]{2} {\color{red}{\frac{u^{\frac{1}{3} + 1}}{\frac{1}{3} + 1}}}=5 \sqrt[3]{2} {\color{red}{\left(\frac{3 u^{\frac{4}{3}}}{4}\right)}}$$

Recorde que $$$u=x + 2$$$:

$$\frac{15 \sqrt[3]{2} {\color{red}{u}}^{\frac{4}{3}}}{4} = \frac{15 \sqrt[3]{2} {\color{red}{\left(x + 2\right)}}^{\frac{4}{3}}}{4}$$

Portanto,

$$\int{5 \sqrt[3]{2 x + 4} d x} = \frac{15 \sqrt[3]{2} \left(x + 2\right)^{\frac{4}{3}}}{4}$$

Adicione a constante de integração:

$$\int{5 \sqrt[3]{2 x + 4} d x} = \frac{15 \sqrt[3]{2} \left(x + 2\right)^{\frac{4}{3}}}{4}+C$$

$$$\int 5 \sqrt[3]{2 x + 4}\, dx = \frac{15 \sqrt[3]{2} \left(x + 2\right)^{\frac{4}{3}}}{4} + C$$$A