Dtermine whether the integral is convergent or divergent, using the Comparison Theorem

Using the Comparison Theorem to determine whether the integral is convergent or divergent. I don't know how to change the form.

$$\int_1^\infty \frac{1}{\sqrt{x^3-0.1}}dx$$

Here, I tried to find what's higher or lower and all attempts were futile. enter image description here

I couldn't find other formations anymore. I want to receive some tips to solve this problem.

Hint: $$\frac1{\sqrt{{x^3-0.1}}}\leq c\frac 1{\sqrt{x^3}}$$ for some $$c>0$$ and sufficiently large $$x$$.
• @user335567 $2(x^3-0.1)\geq x^3$ when $x>1$ – Sui Sep 25 '20 at 7:09
• @user335567 $2(x^3-0.1)= x^3+(x^3-0.2)\geq x^3+0.8$ Are you in middle school? – Sui Sep 25 '20 at 9:09