# Improper integral: $\int_{-1}^{\infty} \frac{dx}{x^2 + \sqrt[3]{x^4 + 1}}$

Decide if the integral

$$\int_{-1}^{\infty} \frac{dx}{x^2 + \sqrt[3]{x^4 + 1}}$$

converges.

I decided to write $\int_{-1}^{\infty} \frac{dx}{x^2 + \sqrt[3]{x^4 + 1}}$ = $\int_{-1}^{1} \frac{dx}{x^2 + \sqrt[3]{x^4 + 1}}$ + $\int_{1}^{\infty} \frac{dx}{x^2 + \sqrt[3]{x^4 + 1}}$

It is easy to show that the second integral converges using inequality. But about the first one, how can I argue that it does converges? My attempt was to finding a function bigger than this one that has an easy integral to calculate (if the function is continuous at the interval, I can just take a constant one that equals to 1+ the local maximum!). But I was wondering: Is that really necessary?

When $x\to \infty$ then $\frac{1}{x^2+\sqrt[3]{1+x^4}}\sim\frac{1}{x^2}$
$$x^2+(x^4+1)^\frac{1}{3}=0$$ does not have a root for any $x$.