How would we find $ \int \sqrt[x]{x}$ or $\int \left( x^{x^{x^{.^{....... \infty}}}} \right) dx$? I know that the antiderivate of $$x^{\frac{1}{x}}$$ cannot be expressed in the form of elementary functions, but even so how will we express it in the form of a power series ?
Inspiration of the question:
Well i was originally trying to find the integral of: 
$$\int \left( x^{x^{x^{.^{...}}}} \right) dx$$
I started of by taking $$ y=x^{x^{x^{.^{...}}}} $$
Therefore, 
$$ y=x^y $$
Taking ln on both sides:
$$ \ln y= y (\ln x) $$
$$ \ln x= \frac {\ln y}{y} $$
$$ \ x= e^ {\frac {\ln y}{y}} $$
$$ e^ {\frac {\ln y}{y}}= y^{\frac{1}{y}} $$
So, 
$$\int \left( x^{x^{x^{.^{...}}}} \right) dx=\int \sqrt[y]{y}dy$$
 A: So finally with some of mrtaurho's help I finally have the answer.
We can write $ x ^ \frac {1}{x} $ as:
$$ x ^ \frac {1}{x}= e ^ { \frac {ln x} {x}} $$
The expansion of $e^x$ is:
$$ e^x=\sum_{n=0}^\infty \frac {x^n}{n!}$$ 
Similarly the expansion of $e ^ { \frac {ln x} {x}}$ is:
$$ e ^ { \frac {ln x} {x}}=\sum_{n=0}^\infty \frac {(\ln x)^n}{x^n.n!} $$
Now evaluating $\int \frac{\log^n(x)}{x^n}\mathrm dx$:
$$\int \frac{\log^n(x)}{x^n}\mathrm dx=\int \frac{\log^n(x)}{x^{n-1}}\frac{\mathrm dx}x\stackrel{\log(x)=t}=\int \frac{t^n}{e^{(n-1)t}}\mathrm dt =\int t^ne^{-(n-1)t}\mathrm dt$$
The structure of the last integral is clearly recognizable as the one given by the Gamma Function. Thus, we can further obtain
$$\int t^ne^{-(n-1)t}\mathrm dt\stackrel{(n-1)t\mapsto t}=\int \left(\frac t{n-1}\right)^ne^{-t}\frac{\mathrm dt}{n-1}=\frac1{(n-1)^{n+1}}\int t^ne^{-t}\mathrm dt$$
Hence the integrand vanishes while $t$ is approaching towards infinity we may write the latter integral as
$$\int (t')^ne^{-t'}\mathrm dt'=-\int_t^\infty (t')^ne^{-t'}\mathrm dt'=-\Gamma(n+1,t)$$
$$\int \frac{\log^n(x)}{x^n}\mathrm dx~=~-\frac1{(n-1)^{n+1}}\Gamma(n+1,(n-1)\log(x))$$
$$\therefore~ \int \sum_{n=0}^\infty \frac {(\ln x)^n}{x^n.n!} dx = \sum_{n=0}^\infty \frac {\Gamma(n+1,(n-1)\log(x))}{(n-1)^{n+1}.n!} = \sum_{n=0}^\infty \frac {\ln ^{n+1}(x) \left(-E_{-n} \left( \left( n-1 \right) \ln x \right) \right)}{n!} $$

$$ \therefore~ \int \left( x^{x^{x^{.^{....... \infty}}}} \right) dx = \int \sqrt[x]{x}dx = \sum_{n=0}^\infty \frac {\ln ^{n+1}(x) \left(-E_{-n} \left( \left( n-1 \right) \ln x \right) \right)}{n!} $$

I would like to name this function "The Anti-Sophomore's Dream" function xD 
