"Suppose $U$ is measurable and defined on $\mathbb{R_{+}}$. Integrate $b(x)/x$" - how to start?

Question

I have part of a proof that I'm struggling with in a textbook where: $$b(x) := \dfrac{xU(x)}{\int_0^x \ U(t)dt}$$ The author suggests that 'integrating $\frac{b(x)}{x}$ leads to the representation': $$\int_0^x \ U(s)ds = c\exp\bigg(\int_1^x \ t^{-1}b(t)dt\bigg)$$ However, I cannot quite see how this occurs. Is the author using the integrating factor somewhere to get this exponential function?

Answer 1

Hint:

If we denote $g(x) = \int_{0}^{x}U(t)dt$, then $g'(x) = U(x)$ (as a derivative of the integral of upper bound). So in terms of $g(x)$, we will have $\frac{b(x)}{x} = \frac{g'(x)}{g(x)}$. So integrating $\frac{b(t)}{t}= t^{-1}b(t)$ is the same as integrating $\frac{g'(t)}{g(t)}$. For which the antiderivative is $\ln (g(t))$.

However I'm not sure with the lower bound 1. Do we know that $x\geq1$?

And I'm sorry. I just didn't manage to fit this in the comments.