Integral of $\sqrt[3]{x}$

Find $\int \sqrt[3]{x}\, dx$.

Apply the power rule $\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$ $\left(n \neq -1 \right)$ with $n=\frac{1}{3}$:

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

Therefore,

$$\int{\sqrt[3]{x} d x} = \frac{3 x^{\frac{4}{3}}}{4}$$

Add the constant of integration:

$$\int{\sqrt[3]{x} d x} = \frac{3 x^{\frac{4}{3}}}{4}+C$$

Answer: $\int{\sqrt[3]{x} d x}=\frac{3 x^{\frac{4}{3}}}{4}+C$