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Integral of $$$\frac{1}{\sqrt[3]{x}}$$$
Related calculator: Definite and Improper Integral Calculator
Your Input
Find $$$\int \frac{1}{\sqrt[3]{x}}\, dx$$$.
Solution
Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=- \frac{1}{3}$$$:
$${\color{red}{\int{\frac{1}{\sqrt[3]{x}} d x}}}={\color{red}{\int{x^{- \frac{1}{3}} d x}}}={\color{red}{\frac{x^{- \frac{1}{3} + 1}}{- \frac{1}{3} + 1}}}={\color{red}{\left(\frac{3 x^{\frac{2}{3}}}{2}\right)}}$$
Therefore,
$$\int{\frac{1}{\sqrt[3]{x}} d x} = \frac{3 x^{\frac{2}{3}}}{2}$$
Add the constant of integration:
$$\int{\frac{1}{\sqrt[3]{x}} d x} = \frac{3 x^{\frac{2}{3}}}{2}+C$$
Answer
$$$\int \frac{1}{\sqrt[3]{x}}\, dx = \frac{3 x^{\frac{2}{3}}}{2} + C$$$A