Answer
$$3\left(x^{2/3} e^{\sqrt[3]{x}}-2 \sqrt[3]{x}e^{\sqrt[3]{x}}+e^{\sqrt[3]{x}} \right)+c$$
Work Step by Step
Given $$\int e^{\sqrt[3]{x}} d x$$
Let $u^3=x\ \Rightarrow \ \ 3u^2 du=dx$, then
$$ \int e^{\sqrt[3]{x}} d x=3\int u^2e^{u} d u$$
Integrate by parts
\begin{align*}
U&=u^2\ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=e^udu\\
dU&=2udu\ \ \ \ \ \ \ \ \ \ \ \ \ dv= e^u
\end{align*}
\begin{align*}
\int e^{\sqrt[3]{x}} d x&=3\int u^2e^{u} d u\\
&=3\left(u^2 e^u-2\int ue^udu \right)
\end{align*}
To compute $ \displaystyle\int ue^udu $
\begin{align*}
U&=u \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=e^udu\\
dU&= du\ \ \ \ \ \ \ \ \ \ \ \ \ dv= e^u
\end{align*}
\begin{align*}
\int ue^udu &=ue^u-\int e^udu\\
&=ue^u-e^u+c
\end{align*}
Hence
\begin{align*}
\int e^{\sqrt[3]{x}} d x&=3\int u^2e^{u} d u\\
&=3\left(u^2 e^u-2 ( ue^u-e^u) \right)+c\\
&=3\left(u^2 e^u-2 ue^u+e^u \right)+c\\
&=3\left(x^{2/3} e^{\sqrt[3]{x}}-2 \sqrt[3]{x}e^{\sqrt[3]{x}}+e^{\sqrt[3]{x}} \right)+c
\end{align*}
