Solve ∫ sqrt[3]{3t}dt

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

Evaluate

$$\frac{3\sqrt[3]{3}t^{\frac{4}{3}}}{4}+С$$

Solution Steps

Factor out the constant using \int af\left(t\right)\mathrm{d}t=a\int f\left(t\right)\mathrm{d}t.

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

Rewrite \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{4}{3}}}{\frac{4}{3}}. Simplify.

$$\sqrt[3]{3}\times \frac{3t^{\frac{4}{3}}}{4}$$

Simplify.

$$\frac{3\sqrt[3]{3}t^{\frac{4}{3}}}{4}$$

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.

$$\frac{3\sqrt[3]{3}t^{\frac{4}{3}}}{4}+С$$

Differentiate w.r.t. t

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