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