How can we solve the integral $\int_\epsilon^1 e^{x^n\log x }\mathrm{d}x$? Set an $\epsilon\in(0,1)$. How can we solve the integral $\int_\epsilon^1 e^{x^n\log x }{\rm d }x$ for all $n\in\mathbb{N}$? 
My attempt.
Set $I_n=\int_\epsilon^1 e^{x^n\log x }{\rm d }x$. I tried a recursive procedure to express $I_{n}$ in terms of $I_{n-1}$:
\begin{align}
I_n=& \int_\epsilon^1 e^{x^n\log x }{\rm d }x\\
   =& \int_\epsilon^1 D_x(x)e^{x^n\log x }{\rm d }x\\
   =& xe^{x^n\log x }\Big|_\epsilon^1-\int_\epsilon^1 xD_x(e^{x^n\log x }){\rm d }x\\
\end{align}
But it did not work. So I tried to use some feature of symbolic integration: (see this answer).

This answer is based on the comments of Lucian and especially Winther.
Expand 
$$
e^{(x^n \log  x)} = \sum_{k=0}^\infty \frac{(x^{n})^k(\log  x)^k}{k!}
$$
Therefore we have 
$$
\int_\epsilon^1 e^{(x^n \log  x)}\,\mathrm{d}x =  \int_\epsilon^1 \sum_{n=0}^\infty \frac{(x^{n})^k(\log  x)^k}{k!} \,\mathrm{d}x 
$$
By uniform convergence of the power series, we may interchange summation and integration
$$
\int_\epsilon^1 e^{(x^n \log  x)}\,\mathrm{d}x = \sum_{n=0}^\infty \int_\epsilon^1 \frac{x^{nk}(\log  x)^k}{k!}\,\mathrm{d}x.
$$
According to Wikpedia
  $\int x^m (\ln x)^k\,\mathrm{d}x
= \frac{x^{m+1}}{m+1}
 \cdot \sum_{i=0}^k (-1)^i \frac{(k)_i}{(m+1)^i} (\ln x)^{k-i}$ for all $m\neq -1$ where $(k)_i$ denotes the falling factorial. Set $m=kn$. Then
\begin{align}
\int_\epsilon^1 \frac{x^{kn}(\ln x)^k}{k!}\,\mathrm{d}x
=& \sum_{i=0}^k (-1)^i
\frac{1}{k!}\cdot\frac{x^{kn+1}}{(km+1)}
 \cdot  \frac{(k)_i}{(km+1)^i} (\ln x)^{k-i}\Bigg|_\epsilon^1
\\
=&\sum_{i=0}^k (-1)^{i+1}
\frac{1}{k!}\cdot\frac{x^{kn+1}}{(km+1)}
 \cdot  \frac{(k)_i}{(km+1)^i} (\ln \epsilon)^{k-i}
\end{align}
and finally
$$
\int_\epsilon^1 e^{x^n\log(x)}\,\mathrm{d}x
= 
\sum_{k=0}^\infty
\sum_{i=0}^k (-1)^{i+1}
\frac{1}{k!}
\cdot
\frac{x^{kn+1}}{(km+1)}
\cdot
\frac{(k)_i}{(km+1)^i} (\ln \epsilon)^{k-i}
$$
