Answer
The integral $\int_{0}^{\infty} \dfrac{dx}{(8x^2+x^4)^{1/3}}$ converges.
Work Step by Step
We are given the function
$f(x)=\int_{0}^{\infty} \dfrac{dx}{(8x^2+x^4)^{1/3}}$
Since, $(8x^2+x^4)^{1/3} \geq (8x^2)^{1/3}$
This yields:
$\dfrac{1}{(8x^2+x^4)^{1/3}} \leq \dfrac{1}{2x^{2/3}} $
But the integral $\int_{0}^{1} \dfrac{dx}{2x^{2/3}}= [\dfrac{3x^{1/3}}{2}]_0^1=\dfrac{3}{2}$
Thus, the integral $\int_{0}^{1} \dfrac{dx}{2x^{2/3}}$ converges. Therefore, by the comparison test, the integral $\int_{0}^{\infty} \dfrac{dx}{(8x^2+x^4)^{1/3}}$ converges as well.
