Using the comparison test I need to determine if the integral is convergent or divergent $$\int_1^{\infty} \frac{1}{\sqrt{x^{3} +1}}dx$$

From what i know about the comparison theorem I need to get the integral in the form of $$\frac{1}{x^{p}} $$ p > 1 (converges) p<= 1 (diverges). So as x gets larger and larger in the integral the "+1" becomes less and less relevant. so can the function be evaluated as $$\int_1^{\infty} \frac{1}{\sqrt{x^{3}}}dx$$ and even as x goes to infinity the square root would become less relevant, so can I just evaluate ? $$\int_1^{\infty} \frac{1}{x^{3}}dx$$ (it's easy to see that the function converges, but i need to use/understand the comparison theorem)

  • 1
    $\begingroup$ Rather, rewrite it as $\frac1{\sqrt{x^3}}=x^{-3/2}$ and use power rule. $\endgroup$ Jun 7, 2017 at 15:16
  • $\begingroup$ Related: math.stackexchange.com/questions/1631639/… $\endgroup$ Jun 7, 2017 at 18:46
  • $\begingroup$ Omitting the square root in the last step is not correct. The +1 is a constant, but $x$ and $\sqrt{x}$ grow at very different rates, even though they both diverge. $\endgroup$
    – Ramanujan
    Dec 26, 2019 at 10:45

2 Answers 2


You are almost right, but you can formalize your argument.

The comparison test can be used this way:

$$\frac 1{\sqrt{x^3+1}}\leqslant \frac 1{\sqrt{x^3}}=\frac 1{x^{3/2}}.$$

And you know that

$$\int_1^\infty \frac 1{x^p}\mathrm d x$$

converges if, and only if, $p>1$.

Since $\frac 32>1$, you can conclude that this integral converges.


$$\int\frac{dx}{\sqrt{x^3+1}} \leq \int\frac{dx}{x^\frac{3}{2}}$$

Edit: You can't just "ignore" the square root. For example if you had $$\int \frac{dx}{\sqrt{x^p}}$$

for $1<p<2$. Then this integral doesn't converge. However, the integral converges if you "ignore" the squre root.


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