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Differentiate w.r.t. z
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\frac{3z^{\frac{4}{3}}}{4}
Rewrite \sqrt[3]{z} as z^{\frac{1}{3}}. Since \int z^{k}\mathrm{d}z=\frac{z^{k+1}}{k+1} for k\neq -1, replace \int z^{\frac{1}{3}}\mathrm{d}z with \frac{z^{\frac{4}{3}}}{\frac{4}{3}}. Simplify.
\frac{3z^{\frac{4}{3}}}{4}+С
If F\left(z\right) is an antiderivative of f\left(z\right), then the set of all antiderivatives of f\left(z\right) is given by F\left(z\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.