Comparing two almost identical improper integrals

Question

I saw a similar problem in my calculus textbook, which got me thinking. I can see why $\int_{-\infty}^01/x^3$ doesn't converge, but how come a very similar function, $\int_{-\infty}^01/\sqrt (x^6+1)$ does converge?

Answer 1

that's the same as integral from 0 to plus infinity. $1/x^3$ is okay around infinity (for big x's) and the $+1$ factor makes it integrable around 0, so the whole integral converges

Answer 2

Because, unlike $\dfrac{1}{x^3}$, the function $\dfrac{1}{\sqrt{x^6+1}}$ behaves nicely near $0$. Indeed, the second function is continuous in the interval $[-1,0]$.