Đặt $u=x+\sqrt{1+{{x}^{2}}}$ thì $u-x=\sqrt{1+{{x}^{2}}}\Rightarrow {{x}^{2}}-2ux+{{u}^{2}}=1+{{x}^{2}}$

$\Rightarrow x=\frac{{{u}^{2}}-1}{2u}\Rightarrow dx=\frac{1}{2}\left( 1+\frac{1}{{{u}^{2}}}du \right)$

Đổi cận $x=-1$ thì $u=\sqrt{2}-1,x=1$ thì $u=\sqrt{2}+1$

$\Rightarrow I=\int_{\sqrt{2}-1}^{\sqrt{2}+1}{\frac{\frac{1}{2}\left( 1+\frac{1}{{{u}^{2}}} \right)du}{1+u}}=\frac{1}{2}\int_{\sqrt{2}-1}^{\sqrt{2}+1}{\frac{du}{1+u}}+\frac{1}{2}\int_{\sqrt{2}-1}^{\sqrt{2}+1}{\frac{du}{\left( 1+u \right){{u}^{2}}}}$

$=\frac{1}{2}\int_{\sqrt{2}-1}^{\sqrt{2}+1}{\frac{du}{1+u}}+\frac{1}{2}\int_{\sqrt{2}-1}^{\sqrt{2}+1}{\left( \frac{1}{{{u}^{2}}}-\frac{1}{u}+\frac{1}{u+1} \right)du=1\Rightarrow a=1}$

$S={{i}^{2016}}+{{i}^{2000}}={{\left( {{i}^{2}} \right)}^{1008}}+{{\left( {{i}^{2}} \right)}^{1000}}={{\left( -1 \right)}^{1008}}+{{\left( -1 \right)}^{1000}}=2$