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)
Best Answer
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.