On $\int \sqrt[x]x$dx. Solution found for $eW\left(\frac1e\right)\le x\le e$. Notice:
Note that there is hope for a more general answer because an answer for the area under the infinite tetration/power tower curve was almost certainly found. It uses the nice OEIS A008405 find. Combined with the Integral of an Inverse function theorem, we can gently extend the interval of convergence via Taylor and Power series:

Integral over domain of infinite tetration of x over extended domain from 0 to $\sqrt[e]e$. Possible $\int_{e^{-e}}^{e^\frac1e} x^{x^{…}}dx$ solution.

Here is a related problem which has a non-integral solution seemingly for any integration bounds. This is different from this question as this addresses an actual solution and hopes of finding a better one. This should be an exact solution. I have also found a solution to the integral of the $x^{th}$ root of $x$ using this source with

all context and work here!

Here are some nice results. This uses the Regularized Incomplete Gamma function, Product Logarithm/W-Lambert function, and Generalized Exponential Integral function. More identities can be verified and derived using the links, but three forms are enough:
$\begin{align*}
\int \sqrt[x]x \, dx &=x^{\frac 1x+1}+\sum_{n\ge 1}(-1)^n n^{n-2}Q\left(n,-\frac{\ln(x)}x\right)+C\\ 
& = x^{\frac1x +1}+\sum_{n\ge 1}\frac{n^{n-1}\ln^n(x)}{n!x^n}E_{1-n}\left(-\frac{\ln(x)}{x}\right)+C\\ 
&= x^\frac1x\left(x+\sum_{n\ge1}\sum_{k=0}^{n-1}\frac{(-1)^{n+k}n^{n-2}\ln^k(x)}{k!x^k}\right),\end{align*}$
for $x\in [e\,W(e^{-1}),e]$.
\begin{align*}
\int_{eW\left(\frac 1e\right)}^e x^\frac 1x \, dx &= e^{1+\frac1e}-e^{1-\frac1e} W\left(\frac 1e\right)-\sum_{n=1}^\infty\frac{(-n)^{n-1}}{n}Q\left(n,-\frac 1e, \frac 1e\right)\\&= 2.516687\ldots
\end{align*}
Source
\begin{align*}
\int_1^e x^\frac1x \, dx &=e^{\frac1e +1}-1+\sum_{n\ge1}(-1)^n n^{n-2}\left(Q\left(n,-\frac1e\right)-1\right)\\
&=2.308328\ldots
\end{align*}
Source
\begin{align*}
\int_1^2 x^\frac1x \, dx & =2\sqrt2-1+\sum_{n\ge 1}(-1)^nn^{n-2}\left (Q\left(n,-\frac{\ln(2)}2\right)-1\right)\\
& =1.277065\ldots
\end{align*}
Source
\begin{align*}
\int_{3W\left(\frac13\right)}^\frac32 x^\frac 1x \, dx &= \left(\frac32\right) ^{\frac53}-{e^{-W\left( \frac13\right)-\frac 13}}+\sum_{n\ge1}(-1)^n n^{n-2} Q\left(n,-\frac23 \ln\left(\frac32\right),\frac 13\right)\\
&=0.788184\ldots
\end{align*}
Source
I received a suggestion for a series of $e^{\frac{\ln(x+1)}{x+1}}$ to “shift” the integration bounds left vertically by 1. I found this series based on this

answer from @mathphile

which does not give the right result as the n=0 term diverges and even trying to use the lower integration bound as $e^{-e}$ still gives the wrong result. The user’s answer would have cracked the question.
The question here is how to find an integral representation without any trivial solutions like those in paragraph 1. What is a general solution to the antiderivative of $\sqrt[x]x$ as the restriction makes the formula here diverge? A solution to $\int_0^{eW\left(\frac1e\right)} \sqrt[x]x \, dx$ is wanted too. Please click the context and work link for the derivation of the formula and other theorems used.
 A: So we have to solve the integral-
$\textstyle\displaystyle{\int_{0}^{e\operatorname{W}\left(\frac{1}{e}\right)}x^{-x}dx}$
Let's just do the indefinite integral first, let
$\textstyle\displaystyle{I=\int x^{-x}dx}$
Substitute, $u=x^{-x}$, then $x=e^{\operatorname{W}(-\ln(u))}$ and $\textstyle\displaystyle{dx=\frac{-du}{u(1+\operatorname{W}(-ln(u)))}}$
$\textstyle\displaystyle{I=-\int\frac{1}{1-(-\operatorname{W}(-\ln(u)))}du}$
After the substitution the lower bound will be $1$ and the upper bound will be $\left(e\operatorname{W}\left(\frac{1}{e}\right)\right)^{-e\operatorname{W}\left(\frac{1}{e}\right)}$
$\left|-\operatorname{W}(-\ln(1))\right|=0<1$
$\left|-\operatorname{W}(-\ln(\left(e\operatorname{W}\left(\frac{1}{e}\right)\right)^{-e\operatorname{W}\left(\frac{1}{e}\right)}))\right|=\operatorname{W}\left(\frac{1}{e}\right)=0.2784...<1$
So we can use the geometric series giving us-
$\textstyle\displaystyle{I=\sum_{n=0}^{\infty}(-1)^{n+1}\int\operatorname{W}^n(-\ln(u))du}$
Let's do the integral in the sum-
$\textstyle\displaystyle{\int\operatorname{W}^n(-\ln(u))du}$
Substitute, $v=-\ln(u)$ then $u=e^{-v}$ and $du=-e^{-v}dv$ giving us-
$\textstyle\displaystyle{-\int e^{-v}\operatorname{W}^n(v)dv}$
Substitute, $t=\operatorname{W}(v)$ then $v=te^t$ and $dv=(t+1)e^tdt$ giving us-
$\textstyle\displaystyle{-\int t^n(t+1)e^{t(1-e^t)}dt}$
$\textstyle\displaystyle{=- \sum_{k=0}^{\infty}\frac{1}{k!}\left(\int t^{n+k+1}(1-e^t)^k dt+\int t^{n+k}(1-e^t)^kdt\right)}$
Using the binomial theorem we have-
$\textstyle\displaystyle{\sum_{k=0}^{\infty}\sum_{j=0}^{k}\frac{(-1)^{j+1}}{k!}{k\choose j}\left(\int t^{n+k+1}e^{jt}dt-\int t^{n+k}e^{jt}dt\right)}$
Finally Substitute, $-w=jt$ then $\textstyle\displaystyle{t=-\frac{w}{j}}$ and $\textstyle\displaystyle{dt=-\frac{1}{j}dw}$ giving us-
$\textstyle\displaystyle{\sum_{k=0}^{\infty}\sum_{j=0}^{k}\frac{(-1)^{j+1}}{j!(k-j)!}\left(\left(-\frac{1}{j}\right)^{n+k+2}\int w^{n+k+1}e^{-w}dw-\left(-\frac{1}{j}\right)^{n+k+1}\int w^{n+k}e^{-w}dw\right)}$
$\textstyle\displaystyle{=\sum_{k=0}^{\infty}\sum_{j=0}^{k}\frac{(-1)^{n+j+k}}{j^{n+k+1}j!(k-j)!}\left(\frac{1}{j}\Gamma(n+k+2,w)+\Gamma(n+k+1,w)\right)}$
So,
$\textstyle\displaystyle{I=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\sum_{j=0}^{k}\frac{(-1)^{j+k+1}}{j^{n+k+1}j!(k-j)!}\left(\frac{1}{j}\Gamma(n+k+2,-j\operatorname{W}(x\ln(x)))+\Gamma(n+k+1,-j\operatorname{W}(x\ln(x)))\right)}$
So finally our definite integral will be-
$\textstyle\displaystyle{\int_{0}^{e\operatorname{W}\left(\frac{1}{e}\right)}x^{-x}dx}$
$\textstyle\displaystyle{=\bigg[\sum_{n,k\geq 0}\sum_{j=0}^{k}\frac{(-1)^{j+k+1}}{j^{n+k+1}j!(k-j)!}\left(\frac{1}{j}\Gamma(n+k+2,-j\operatorname{W}(x\ln(x)))+\Gamma(n+k+1,-j\operatorname{W}(x\ln(x)))\right)\bigg]_{0}^{e\operatorname{W}\left(\frac{1}{e}\right)}}$
Notice that,
■$\textstyle\displaystyle{\lim_{x\rightarrow 0}\operatorname{W}(x\ln(x))=0}$
■$\textstyle\displaystyle{\operatorname{W}\left(e\operatorname{W}\left(\frac{1}{e}\right)\ln\left(e\operatorname{W}\left(\frac{1}{e}\right)\right)\right)=-\operatorname{W}\left(\frac{1}{e}\right)}$
So we have,
$\textstyle\displaystyle{\bigg[\sum_{n,k\geq 0}\sum_{j=0}^{k}\frac{(-1)^{j+k+1}}{j^{n+k+1}j!(k-j)!}\left(\frac{1}{j}\Gamma\left(n+k+2,j\operatorname{W}\left(\frac{1}{e}\right)\right)+\Gamma\left(n+k+1,j\operatorname{W}\left(\frac{1}{e}\right)\right)\right)\bigg]-\bigg[\sum_{n,k\geq 0}\sum_{j=0}^{k}\frac{(-1)^{j+k+1}}{j^{n+k+1}j!(k-j)!}\left(\frac{1}{j}\Gamma(n+k+2,0)+\Gamma(n+k+1,0)\right)\bigg]}$
$\textstyle\displaystyle{=\sum_{n,k\geq 0}\sum_{j=0}^{k}\frac{(-1)^{j+k+1}}{j^{n+k+1}j!(k-j)!}\left(\frac{1}{j}\Gamma\left(n+k+2,j\operatorname{W}\left(\frac{1}{e}\right),0\right)+\Gamma\left(n+k+1,j\operatorname{W}\left(\frac{1}{e}\right),0\right)\right)}$
$\textstyle\displaystyle{=\sum_{n,k\geq 0}\sum_{j=0}^{k}\frac{(-1)^{j+k}}{j^{n+k+1}j!(k-j)!}\left(\frac{1}{j}\gamma\left(n+k+2,j\operatorname{W}\left(\frac{1}{e}\right)\right)+\gamma\left(n+k+1,j\operatorname{W}\left(\frac{1}{e}\right)\right)\right)}$
We can easily get a recurrence relation for the lower incomplete gamma function by integration by parts-
$\gamma(n+1,a)=-a^ne^{-a}+n\gamma(n,a)$
Giving us,
$\textstyle\displaystyle{=\sum_{n,k\geq 0}\sum_{j=0}^{k}\frac{(-1)^{j+k}}{j!(k-j)!}\left(\frac{n+k+2}{j^{n+k+1}}\gamma\left(n+k+1,j\operatorname{W}\left(\frac{1}{e}\right)\right)-\frac{1}{j}\operatorname{W}^{n+k}\left(\frac{1}{e}\right)e^{-1-(j+1)\operatorname{W}\left(\frac{1}{e}\right)}\right)}$
$\boxed{\textstyle\displaystyle{\int_{0}^{e\operatorname{W}\left(\frac{1}{e}\right)}x^{-x}dx=\sum_{n,k\geq 0}\sum_{j=0}^{k}\frac{(-1)^{j+k}}{j!(k-j)!}\left(\frac{n+k+2}{j^{n+k+1}}\gamma\left(n+k+1,j\operatorname{W}\left(\frac{1}{e}\right)\right)-\frac{1}{j}\operatorname{W}^{n+k}\left(\frac{1}{e}\right)e^{-1-(j+1)\operatorname{W}\left(\frac{1}{e}\right)}\right)}}$
I can't simplify more than this, the series representation is kind of useless but this all I can give you. And please tell me if there any mistakes.
