Let $f: [0, 1] \rightarrow \mathbb{R}$ be a continuous function.
Suppose that $f(0) = 0, \forall x \in (0, 1]: f(x) > 0$ and the right-hand derivative $f'_{+}(0)$ exists.
I'm trying to prove that the improper integral $\int_{0}^{1}\frac{1}{f(x)} dx$ does not exist. I struggle to interpret the assumptions, I took some examples but I'm not sure how I can relate $\frac{1}{f}$ to $f$ directly.
I'd like to have some guidance, thanks
Let $a:=f'_+(0)$ and $b:=|a|+1$ (actually, $a\ge0$ but we shall not use it).
By the assumptions, $$\forall x\in(0,1]\quad 0<f(x)=x\left(a+\varepsilon(x)\right)$$ for some function $\varepsilon$ such that $\lim\limits_{x\to0^+}\varepsilon(x)=0.$
Therefore, for $x>0$ close enough to $0,$ $$0<f(x)\le bx$$ hence $$\frac1{f(x)}\ge\frac1{bx},$$ which makes your improper integral equal to $+\infty.$