Evaluating $\int{e^{x^{1/3}}dx}$ How can I get  $$\int{e^{x^{1/3}}dx}$$
I think integrating by parts may work, but I can't figure out the exact way.
 A: $$\int e^{x^{1/3}} dx \stackrel{x=t^3}{=} 3\int t^2 e^t dt \stackrel{(*)}{=} 3 (t^2 - 2t + 2) e^t + C \stackrel{t=x^\frac{1}{3}}= 2(x^\frac{2}{3} - 2x^\frac{1}{3} + 2) e^{x^\frac{1}{3}} + C$$
$(*)$: We see $$(t^2 e^t)' = (t^2 + 2t) e^t$$
And iterate
$$\begin{align*}
((t^2 - 2t)e^t)' & = (t^2 + 2t - 2t - 2)e^t \\
((t^2 - 2t + 2) e^t)' & = (t^2 + 2t - 2t - 2 + 2)e^t = t^2 e^t
\end{align*}$$
A: another solution : 
$$ x = t^3 \Rightarrow $$
then we have to evaluate 
$$ I = \int t^2 e^t \ dt $$
$$ u = t^2 \quad , dv = e^t \ dt $$
$$ du = 2t \ dt \quad , v = e^t $$
$$ \Rightarrow I = uv - \int v \ du = t^2 e^t - 2\int te^t \ dt $$
again $$ u  = t \quad , dv = e^t \ dt $$
$$ du = dt \quad , v = e^t $$
$$ \Rightarrow I = t^2 e^t - 2\left( te^t - \int e^t \ dt \right) = t^2e^t -  2te^t + e^t + C  $$
if you don't know integrating by parts :
$$ d(uv) = u \ dv + v \ du $$
$$ \Rightarrow \int d(uv) = \int u \ dv + \int v \ du $$
$$ \Rightarrow uv = \int u \ dv + \int v \ du  \Rightarrow \int u \ dv = uv -  \int v \ du $$
:-)
