Limit of integral function for $x\to +\infty$

Question

How can you prove that

$$ \lim_{x\to +\infty}\int_{x}^{2x}\frac{dt}{1+t\log t}=0? $$

I think it is not possible to use Cauchy criterion, as

$$ \int_1^{+\infty}\frac{dt}{1+t\log{t}} $$

diverges. Thank you!

Answer 1

Hint: For $x>1$, squeeze $$ 0<\int_x^{2x}\frac{\mathrm{d}t}{1+t\log t}<\int_x^{2x}\frac{\mathrm{d}t}{t\log t} $$

Answer 2

If $x\geq 10$, the function $x\mapsto \frac{1}{1+x\log(x)}$ is decreasing, and thus $$0\leq \int_x^{2x}\frac{1}{1+t\log(t)}\,\mathrm d t\leq\frac{1}{1+x\log(x)}\int_x^{2x}\,\mathrm d t=\frac{x}{1+x\log(x)}\leq\frac{1}{\log(x)}\underset{x\to \infty }{\longrightarrow }0. $$