[disclaimer: this is slightly more advanced maths than that would appear on a STEP paper - contour integration is second year material at Cambridge, so this is just for fun!]

It's a general result (most easily provable using contour integration that $\displaystyle \int_0^{\infty} \frac{1}{x^{n} + 1} = \frac{\pi}{n\sin \frac{\pi}{n}}$ or alternatively, some double integrals and a handy beta/gamma spot works).

Here we have $$\int_0^{\infty} \frac{x^2}{x^6 + 1} \, \mathrm{d}x + \int_0^{\infty} \frac{1}{x^6 +1}$$ the former can be expressed as $\int_0^{\infty} \frac{\mathrm{d}u}{u^2 + 1} = \frac{1}{3}\arctan u \big]_0^{\infty}= \frac{\pi}{6}$ via $u = x^3$.

So all in all, we get $$\int_0^{\infty} \frac{x^2+1}{x^6 +1 } \, \mathrm{d}x = \frac{\pi}{6} + \frac{\pi}{6\sin \frac{\pi}{6}} = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$$