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This problem is similar to this

Using a similar approach, we conclude that the given integral reduces to:

$\bigg( \displaystyle \sum_{r=0}^{\infty} \frac{(-1)^{r} \cdot (r+1)}{2^{r+1}} \bigg) \cdot 2016 \cdot \frac{1}{2}$

$\displaystyle \Rightarrow 504 \cdot \bigg( \displaystyle \sum_{r=0}^{\infty} \big(\frac{-1}{2}\big)^{r} \cdot (r+1) \bigg)$

Let $S = \displaystyle \sum_{r=0} \big( \frac{-1}{2} \big)^{r} \cdot (r+1)$

$\displaystyle \Rightarrow S = 1 + \frac{(-1)(2)}{2} + \frac{(1)(3)}{2^2} + \ldots$

$\displaystyle \Rightarrow -2S = - 2 + 2 + \frac{(-1)(3)}{2} + \frac{(1)(4)}{2^2} + \ldots$

$S-(-2S) = (1-0) + \big( \frac{(-1)(2)}{2} - \frac{(-1)(3)}{2} \big) + \big( \frac{(1)(3)}{2^2} - \frac{(1)(4)}{2^2} \big) + \ldots$

$\displaystyle \Rightarrow 3S = 1 + \frac{1}{2} + \frac{-1}{4} + \ldots$

$\displaystyle \Rightarrow 3S = 1 + \frac{\frac{1}{2}}{1-\frac{-1}{2}}$

$\displaystyle \Rightarrow S = \frac{4}{9}$

Hence, the given integral now reduces to:

$504 \cdot \frac{4}{9} = \boxed{224}$