Answer
Converges
Work Step by Step
Re-arrange the given integral as follows:
$\int_{0}^{1} \dfrac{1}{ \sqrt {x^6+1}} dx+\int_{1}^{\infty} \dfrac{1}{ \sqrt {x^6+1}} dx$ ....(1)
Here, the integral $\int_{0}^{1} \dfrac{1}{ \sqrt {x^6+1}} dx \leq \int_{0}^{1} \dfrac{1}{ \sqrt {0+1}} dx=1$; converges
and
$\int_{1}^{\infty} \dfrac{1}{ \sqrt {x^6+1}} dx \leq \int_{1}^{\infty} \dfrac{1}{ \sqrt {x^6+0}} dx=\lim\limits_{a \to \infty}[\dfrac{-1}{ 2x^2}]_{1}^{a} $
or,
$\lim\limits_{a \to \infty}[\dfrac{-1}{ 2a^2}] -\lim\limits_{a \to \infty}[\dfrac{-1}{ 2}] =\dfrac{1}{2}$; converges