Answer
Converges.
Work Step by Step
We need to re-write the given integral as follows:
$\int_{0}^{1} \dfrac{1}{ \sqrt {x^6+1}} dx+\int_{1}^{\infty} \dfrac{1}{ \sqrt {x^6+1}} dx$
But $\int_{0}^{1} \dfrac{1}{ \sqrt {x^6+1}} dx \leq \int_{0}^{1} \dfrac{1}{ \sqrt {0+1}} dx=1$
which converges.
And
$\int_{1}^{\infty} \dfrac{1}{ \sqrt {x^6+1}} dx \leq \int_{1}^{\infty} \dfrac{1}{ \sqrt {x^6+0}} dx=\int_1^\infty x^{-3} dx \\=\lim\limits_{a \to \infty}[\dfrac{-1}{ 2x^2}]_{1}^{a} \\=\lim\limits_{a \to \infty}[\dfrac{-1}{ 2a^2}] -\lim\limits_{a \to \infty}[\dfrac{-1}{ 2}] $
$=\dfrac{1}{2}$
Thus, the integral converges.