Solve ∫ dt/sqrt[3]{t}

$$\int \frac { d t } { \sqrt[ 3 ] { t } }$$

Evaluate

$$\frac{3t^{\frac{2}{3}}}{2}+С$$

Solution Steps

Rewrite \frac{1}{\sqrt[3]{t}} as t^{-\frac{1}{3}}. Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{-\frac{1}{3}}\mathrm{d}t with \frac{t^{\frac{2}{3}}}{\frac{2}{3}}. Simplify.

$$\frac{3t^{\frac{2}{3}}}{2}$$

If F\left(t\right) is an antiderivative of f\left(t\right), then the set of all antiderivatives of f\left(t\right) is given by F\left(t\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$\frac{3t^{\frac{2}{3}}}{2}+С$$

Differentiate w.r.t. t

$$\frac{1}{\sqrt[3]{t}}$$