Simplify integral of inverse of derivative. I need to simplify function $g(x)$ which I describe below.
Let $F(y)$ be the inverse of $f'(\cdot)$ i.e. $F = \left( f'\right)^{-1}$ and $f(x): \mathbb{R} \to \mathbb{R}$, then $$g(x) =\int_a^x F(y)dy$$ Is it possible to simplify $g(x)$?

Let $t = F(y)$. Then we get that $y = F^{-1}(t) = f'(t)$. Hence, $dy = f''(t) dt$. Hence, we get that 
\begin{align}
g(x) & = \int_{F(a)}^{F(x)} t f''(t) dt\\
& = \left. \left(t f'(t) - f(t) \right) \right \rvert_{F(a)}^{F(x)}\\
& = F(x) f'(F(x)) - f(F(x)) - (F(a) f'(F(a)) - f(F(a)))\\
& = xF(x) - f(F(x)) - aF(a) + f(F(a))
\end{align}
