\( \int _{}^{}\frac{e^{\sqrt [3]{x}}}{\sqrt [3]{x}} d x \)

To tackle the integral \( \int \frac{e^{\sqrt[3]{x}}}{\sqrt[3]{x}} \, dx \), we can use substitution to simplify our work. Let \( u = \sqrt[3]{x} \), which implies that \( x = u^3 \) and \( dx = 3u^2 \, du \). Substituting these into the integral gives us: \[ \int \frac{e^{u}}{u} \cdot 3u^2 \, du = 3 \int u e^u \, du \] Next, we can apply integration by parts here, where we let \( v = e^u \) and \( w' = u \). This results in \( dv = e^u \, du \) and \( w = \frac{u^2}{2} \). Thus: \[ 3 \left( u e^u - \int e^u \, du \right) = 3 \left( u e^u - e^u \right) + C \] Now reverting back to the variable \( x \) gives us \( u = \sqrt[3]{x} \): \[ 3 \left( \sqrt[3]{x} e^{\sqrt[3]{x}} - e^{\sqrt[3]{x}} \right) + C \] Therefore, the integral evaluates to: \[ 3 e^{\sqrt[3]{x}} \left( \sqrt[3]{x} - 1 \right) + C \]