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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.

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1 Answer 1

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Hint: $\frac1{\sqrt{{x^3-0.1}}}\leq c\frac 1{\sqrt{x^3}}$ for some $c>0$ and sufficiently large $x$.

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  • $\begingroup$ I don't think 1/root x^3-0.1 is lower than c X 1/root x^3. root x^3-0.1 is lower than root x^3, then 1/root x^3-0.1 is higher than c X 1/root x^3. could you give me some more details? $\endgroup$
    – user335567
    Sep 25, 2020 at 6:54
  • $\begingroup$ @user335567 $2(x^3-0.1)\geq x^3$ when $x>1$ $\endgroup$
    – Sui
    Sep 25, 2020 at 7:09
  • $\begingroup$ why do you think that?.. when x>1, then 2(x^3-1)>1.8 and x^3>1 but it doesn't mean that 2(x^3-1) > x^3. it just means it would be possible for 2(x^3-1) to be higher than x^3. Could you explain it more concretely? $\endgroup$
    – user335567
    Sep 25, 2020 at 8:56
  • $\begingroup$ @user335567 $2(x^3-0.1)= x^3+(x^3-0.2)\geq x^3+0.8$ Are you in middle school? $\endgroup$
    – Sui
    Sep 25, 2020 at 9:09
  • $\begingroup$ i'm not. thank you. you're a genius $\endgroup$
    – user335567
    Sep 25, 2020 at 9:58

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