We're asked to prove whether the above integral is convergent or divergent. And I would love to have some feedback on whether my proof is right or wrong, or how can I improve it.
Proof:
Notice that for $x\in{}(0,1]$ we have that $x-\ln(x)>0$ since $\ln(x)\leq0$ for $0<x\leq{}1$. And since $x<\sqrt{x}$ on this interval, it holds that $x-\ln(x)>\sqrt{x}$ such that $\frac{1}{\sqrt{x}}>\frac{1}{x-\ln(x)}>0$.
Now consider $\int_0^{1}\frac{1}{\sqrt{x}}\mathbb{d}x$. This integral converges since $\lim_{c\to0^+}\int_{c}^{1}\frac{1}{\sqrt{x}}\mathbb{d}x=0<\infty$ hence by the comparison lemma, $\int_{0}^{1}\frac{1}{x-\ln(x)}\mathbb{d}x$ converges as well. QED
Thanks in advance