$f:\mathbb{R}^n \to \mathbb{R}$ has expansion $\sum_i g_i(x)x^i$ Problem 2-35 on page 34 of Spivak's Calculus on Manifolds states 

If $f: \mathbb{R}^n: \to \mathbb{R}$ is differentiable and $f(0) =0$,
  prove that there exist $g_i: \mathbb{R}^n \to \mathbb{R}$ such that
  $f(x) = \sum_{i=1}^n x^ig_i(x)$. 
Hint: if $h_x(t) = f(tx)$, then $f(x)
 = \int_0^1 h_x'(t)$.

I understand the answer he wants, but hasn't he left out a hypothesis that would ensure $h_x'(t)$ is integrable? (For instance the continuity of $df$.) 
Secondly, I'm wondering if this theorem is used anywhere. Perhaps in differential geometry it might be useful to write locally $f \in C^\infty (M)$ as $\sum_i g_i(x) x_i$ for some coordinates $x_i$.
 A: You are quite right that you need integrability of the derivative. But the answer is quite simple: on page 28 he writes: First of all, we will be interested almost exclusively in functions $f:\mathbf{R}^n\rightarrow\mathbf{R}^n$ which are $C^\infty$ (that is, each component function $f$ possesses continuous partial derivatives of all orders); sometimes we will use the words "differentiable" or "smooth" to mean $C^\infty$". So when he says "differentiable" it is understood "of class $C^\infty$".
The main application of this result (Hadamard's lemma) I know of is to show that the space of vector fields $\mathfrak{X}(M)$ is canonically isomorphic to the space of derivations $\mathrm{Der}_{\mathbf{R}}(C^\infty(M))$. Variations of the result you mention appear later on in Spivak, when he gives a coordinate free proof of $L_X Y=[X,Y]$.
A: I remember this problem. I'm also working through Calc on Manifolds right now. 
Following the hint, we have:
$ h_{\vec{x}}(t) = f(t\vec{x}) $ implies $ f(\vec{x}) = \displaystyle \int_0^1 h_{\vec{x}}'(t) dt $
Then we may write:
$ f(\vec{x}) = \displaystyle \int_0^1 h_{\vec{x}}'(t) dt = \displaystyle \int_0^1 f'(t\vec{x}) dt $
Where $f'(t\vec{x})$ here denotes an ordinary derivative with respect to $t$. Now, we may expand this expression using the chain rule (I will use Spivak's notation since I assume you are familiar with it):
$\displaystyle \int_0^1 f'(t\vec{x}) dt = \displaystyle \int_0^1 \displaystyle \sum_{j=0}^n x^j D_j f(t\vec{x}) dt $
And finally since we are integrating with respect to $t$:
$\displaystyle \int_0^1 \displaystyle \sum_{j=0}^n x^j D_j f(t\vec{x}) dt =\displaystyle \sum_{j=0}^n x^j\displaystyle \int_0^1 D_j f(t\vec{x}) dt$
Now simply let $ g_j (\vec{x})=\displaystyle \int_0^1 D_j f(t\vec{x}) dt$
And we have: $f(\vec{x}) = \displaystyle \sum_{j=0}^n x^j g_j (\vec{x})$
As desired. I recall having a bit of trouble with this one. Also, as you can see, differentiability of $f$ is sufficient to ensure the existence of such $g_j$. As for your final question, once I get further in the book, I may be able to answer it for you, but as of right now, I cannot.
