Expanding $\left| [\widehat{f}( \xi + h) - \widehat{f}( \xi)]/h - [- 2 \pi i f(x)]^{\wedge}(\xi) \right|$ into one integral

Question

In the context of Fourier Analysis, I'm having trouble why, as a lemma to another proof, that

$$ \left| [\widehat{f}( \xi + h) - \widehat{f}( \xi)]/h - [- 2 \pi i x f(x)]^{\wedge}(\xi) \right| = \left| \int_{\mathbb{R}} f(x) e^{- 2 \pi i x \xi} \left[{e^{- 2 \pi i x h} - 1\over h} + 2 \pi i x\right] dx \right| $$

Attempt. I have computed the following:

1. $\widehat{f}(\xi) = \int_{\mathbb{R}} e^{-2 \pi i x \xi} f(x) dx$
2. $\widehat{f}(\xi + h) = \int_{\mathbb{R}} e^{-2 \pi i x (\xi + h)} f(x) dx$
3. $[-2 \pi i x f(x)]^{\wedge}(\xi) = \int_{\mathbb{R}} e^{- 2 \pi i x \xi} \left( -2 \pi i x f(x) \right) dx$

Then

\begin{align*} &\ \ \ \ \ \ \left| [\widehat{f}( \xi + h) - \widehat{f}( \xi)]/h - [- 2 \pi i x f(x)]^{\wedge}(\xi) \right| \\ &= {\int_{\mathbb{R}} e^{-2 \pi i x (\xi + h)} f(x) dx - \int_{\mathbb{R}} e^{-2 \pi i x \xi} f(x) dx \over h} - \int_{\mathbb{R}} e^{- 2 \pi i x \xi} \left( -2 \pi i x f(x) \right) dx \end{align*}

From here it is not clear how to assemble this into the integral above.