Does the function $f$ belong to $L^1_{loc}$ or $L^1$?

Question

I have the following question:

Prove that the function $x \to \dfrac{1}{x(\sqrt{x}-1)} \in L^1_{loc}(]0,+\infty[)$. This function is it in $L^1(]0,+\infty[)$?

I try to do this: first, to prove that $\dfrac{1}{x(\sqrt{x}-1)} \in L^1_{loc}(]0,+\infty[)$, we have that the function is continuous in $]0,+\infty[ / {1}$ so it is $L^1_{loc} (]0,+\infty[ / {1})$. It remains to prove that the function is integrable in the neighbourhood of $1$. So i try to prove that $\displaystyle\int_1^a \dfrac{1}{x(\sqrt{x}-1}= dx < +\infty$. But

1. I have o idea how to do this.

2. How we see il $\dfrac{1}{x(\sqrt{x}-1)} \in L^1(]0,+\infty[)$?

Thank you in advance.

Answer 1

Hint. Note that as $x\to 1$, $$\dfrac{1}{x(\sqrt{x}-1)}=\dfrac{(\sqrt{x}+1)}{x(x-1)}\sim\frac{2}{x-1}.$$ Is this function integrable in a neighbourhood $1$?