Let $$\psi(x)=x-\frac{1-F(x)}{f(x)},$$ where $x \in [\underline{x},\bar{x}]$ and $f(x)$ is a density function.
Proposition: $\;\;\;E(\psi(x))=\underline{x}$
Note that since $x$ is continuous, $$E(\psi(x))=\int_\underline{x}^{\overline x} \psi(x)f(\psi(x))dx.$$
I used integral by parts, but can't remove $F(x)$ part.
How to derive that result?
I think there's a typo in your expectation. I shall also assume $F(x) = \int_{-\infty}^x f(t) \, d t$ is the cumulative distribution function of the density $f(x)$. Note that $F(\underline{x}) = 0$ and $F(\overline{x}) = 1$.
We have $$ E(\psi(x)) = \int_{\underline{x}}^{\overline{x}} \psi(x) f(x) \, d x = \int_{\underline{x}}^{\overline{x}} \Bigl( x - \frac{1 - F(x)}{f(x)} \Bigr) f(x) \, d x. $$ Expanding this we get $$ E(\psi(x)) = \int_{\underline{x}}^{\overline{x}} x f(x) \, d x - \int_{\underline{x}}^{\overline{x}} \, d x + \int_{\underline{x}}^{\overline{x}} F(x) \, d x. $$ The middle integral is easy, it's just $\overline x - \underline x$. The first integral we use integration by parts on, and get $$ \begin{align*} \int_{\underline{x}}^{\overline{x}} x f(x) \, d x &= x F(x) \biggr\rvert_{\underline{x}}^{\overline{x}} - \int_{\underline{x}}^{\overline{x}} F(x) \, d x \\ &= \overline{x} F(\overline{x}) - \underline{x} F(\underline{x}) - \int_{\underline{x}}^{\overline{x}} F(x) \, d x. \\ &= \overline{x} - \int_{\underline{x}}^{\overline{x}} F(x) \, d x\end{align*}$$ since $F(\overline{x}) = 1$ and $F(\underline{x}) = 0$.
Putting all of this together, $$ E(\psi(x)) = \overline{x} - \overline{x} + \underline{x} - \int_{\underline{x}}^{\overline{x}} F(x) \, d x + \int_{\underline{x}}^{\overline{x}} F(x) \, d x = \underline{x}$$ as desired.