Answer
$$
\int e^{\sqrt[3]{x}} d x= 3 e^{\sqrt[3]{x}}\left(x^{2 / 3}-2 x^{1 / 3}+2\right)+C
$$
where $C$ is a constant,
Work Step by Step
$$
\int e^{\sqrt[3]{x}} d x
$$
Let $ w =\sqrt[3]{x} $. Then $ w^{3} =x $ and $3 w^{2} dw =dx $, so
$$
\int e^{\sqrt[3]{x}} d x= 3 \int e^{w} w^{2} d w =3 I
$$
integrate by parts twice, first with
$$
\left[\begin{array}{ll}{u= w^{2} ,} & {d v=e^{w} d w} \\ {d u=2w dw } & {v= e^{w} }\end{array}\right]
$$
so
$$
I = \int e^{w} w^{2} d w =w^{2} e^{w}-\int 2 w e^{w} d w
$$
Now let
$$
\left[\begin{array}{ll}{U=w ,} & {dV=e^{w} d w} \\ {d U=dw } & {V=e^{w} }\end{array}\right].
$$
Thus
$$
\begin{aligned}
I = \int e^{w} w^{2} d w &=w^{2} e^{w}-\int 2 w e^{w} d w \\
&=w^{2} e^{w}-2\left[w e^{w}-\int e^{w} d w\right] \\
&=w^{2} e^{w}-2 w e^{w}+2 e^{w}+C_{1}
\end{aligned}
$$
where $C_1$ is constant, and hence
$$
3 I=3 e^{w}\left(w^{2}-2 w+2\right)+C
$$
Now putting $w =\sqrt[3]{x}$ back again, we get :
$$
\int e^{\sqrt[3]{x}} d x= 3 e^{\sqrt[3]{x}}\left(x^{2 / 3}-2 x^{1 / 3}+2\right)+C
$$
where $C=3C_1$ is a constant