if the indefinite integral of $x^x$ was $f(x)$ what would the indefinite integral of $x^{1/x}$ be in terms of $x$ and $f(x)$? so what this means is if $f(x)$ is the indefinite integral of $x^x dx$ then what would the indefinite integral of $x^{1/x}$ be in terms of $x$ and $f(x)$

Partial answer
Let be $g : x \mapsto \int_1^x t^{1/t} \textrm{d}t$ and $h : x \mapsto \int_0^x t^{-t} \textrm{d}t$.
Thus, if we do the change of variable $u = 1/t$ in the first integral, we have, for all $x \geq 1$:
\begin{align*}
\int_1^x t^{1/t} \textrm{d}t & = -\int_{1/x}^1 u^{-(u + 2)} \textrm{d}u \\
& = -\int_{1/x}^1 1/u^2 u^{-u} \textrm{d}u \\
& = -\left(\left[h(u)/u^2\right]_{1/x}^1 + 2\int_{1/x}^1 \dfrac{h(u)}{u^3} \textrm{d}u\right) \\
& = \left(x^2 h(1/x) - h(1)\right) - 2\int_{1/x}^1 \dfrac{h(u)}{u^3} \textrm{d}u
\end{align*}
We then have:
\begin{equation*}
g(x) = (x^2h(1/x) - h(1)) - 2\int_{1/x}^1 h(u)/u^3 \textrm{d}u
\end{equation*}
We can continue this by 2nd MVT theorem, as $h$ is continuous (even differentiable!) and $u \mapsto 1/u^3$ non negative and continuous, there is some $c_x \in [1/x, 1]$ such that:
\begin{equation*}
\int_{1/x}^1 \dfrac{h(u)}{u^3} \textrm{d}u = -\dfrac{h(c_x)}{2}\left(1 - x^2\right)
\end{equation*}
Finally:
\begin{equation*}
g(x) = x^2 (h(1/x) - h(c_x)) - h(1) + h(c_x)
\end{equation*}
Now, I'm unsure if there is an easy relation to $f : x \mapsto \int_{0}^x t^t \textrm{d}t$, if you try to apply some techniques, you will see that it's difficult to get $-t$.
