For some $n \in N$ and $f: \Bbb R \rightarrow \Bbb R$ the unique differentiable function such that
$$[f(x)]^{2n+1}+f(x)-x=0$$
for all $x$.
For $x \ge 0$ evaluate
$$\int_{0}^{x} f(t)dt$$
P.S: I have tried some ideas like derivation of the functional equation and I got that $f(x)>0$ and $f'(x)>0$. Any idea?
Let $g(x) = x^{2n+1} + x$. Notice that $g$ is the inverse of $f$. Then with the substitution $u = f(t)$,
\begin{align*} \int_{0}^{x} f(t) \, dt &= \int_{0}^{f(x)} u g'(u) \, du \\ &= x f(x) - \int_{0}^{f(x)} g(u) \, du \tag{*} \\ &= x f(x) - \frac{f(x)^{2n+2}}{2n+2} - \frac{f(x)^2}{2}. \end{align*}
Notice that $\text{(*)}$ can be rewritten as a more geometrically suggesting form
$$ \int_{0}^{a} f(x) \, dx + \int_{0}^{f(a)} f^{-1}(y) \, dy = af(a). $$
This can be derived by interpreting the left-hand side as the area of the rectangle with corners $(0,0)$, $(a,0)$, $(a,f(a))$, $(0,f(a))$ which is cut through by the graph of $y = f(x)$.