How to have a form like $y=\int_{}^{}\operatorname{expression }$ where $y=x^5-x+c$

Question

This question is an attempt to study the following polynomials :

Let $0<x<1$ then define

$$y=x^5-x+c$$

I come up with an integral expression which is :

$$y=c+\lim_{n\to \infty}\int_{x}^{x^{5}}\left(y^{\frac{1}{y^{n}}}-1\right)dy\tag{I}$$

Then and it's my question how to have only one integral without the coefficient $c$ on one side starting from $(I)$ ? or :

$$y=\int_{}^{}\operatorname{expression }$$

I have tried to inverse the function : $$f(x)=x^5-x$$ but it involves hypergeometric function .

How to do that ?

Answer 1

Some partial progress :

The main idea is for the use of Fubini's theorem so we introduce a double integral :

$\exists n>0$ fixed and $\exists a,c,d\in(-\infty,\infty)$ such that in the neightborhood of $x,x\in(0,\varepsilon),0<\varepsilon<1$ we have :

$$\int_{x}^{x^{5}}\left(\int_{0}^{1}\left(y^{\frac{1}{y^{n}}}-c\right)dy-\frac{a}{x}\right)dt\simeq x^5-x+d$$

For example :

$$f(x)=f\left(x\right)=\int_{x}^{x^{5}}\left(\int_{0}^{1}\left(y^{\frac{1}{y^{n}}}-c\right)dy-\frac{a}{x}\right)dt$$

Then $0<x<1$:

$$f\left(x-u\right)\simeq x^5-x+d,u=-0.18,n=0.7,c=1.3,a=-0.6,d=0.45$$