Floor on the floor ??

We could break up the limits of the integral, and hence instead of $\int_{-1}^{1}$ it can be $\int_{-1}^{0}$ and $\int_{0}^{1}$

The function $\Bigg \lfloor 1008 + \dfrac{ \lfloor x \rfloor}{x^{2014} + 1}\Bigg \rfloor$ is

$\Bigg \lfloor 1008 + \dfrac {-1}{x^{2014} + 1} \Bigg \rfloor$ for x in the interval $(-1,0)$

Since x belongs to $(-1,0)$ , $\dfrac{-1}{x^{2014} + 1}$ is a small negative quantity.

Hence

$\Bigg \lfloor 1008 + \dfrac {-1}{x^{2014} + 1} \Bigg \rfloor$ for x in the interval $(-1,0)$ is equal to $\boxed{1007}$

and

$\Bigg \lfloor 1008 + \dfrac {0}{x^{2014} + 1} \Bigg \rfloor$ for x in the interval $(0,1)$

Since x belongs to $(0,1)$ , $\dfrac{1}{x^{2014} + 1}$ is a small positive quantity.

Hence

$\Bigg \lfloor 1008 + \dfrac {-1}{x^{2014} + 1} \Bigg \rfloor$ for x in the interval $(0,1)$ is equal to $\boxed{1008}$

Hence the integral is nothing but,

$\boxed{1007 + 1008} = \boxed{2015}$

The year we are in currently.