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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 $g(x)$?

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By "inversion" do you mean the inverse, $(f')^{-1}(x)$, or the reciprocal, $\frac{1}{f'(x)}$? – Arturo Magidin Jun 4 '12 at 5:05
    
thank you for question. I meant the inverse $(f')^{-1}(x)$ – ashim Jun 4 '12 at 5:12

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}

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