Mathematical Analysis I - 4-4-4.(16)

$\begin{aligned} A & = \int_0^4 (x-2) \sqrt{x^2+4x+1}\ dx \\ & = \int_0^4 (x-2) \sqrt{\blue{(x+2)}^2 - 3}\ dx & \small \blue{\text{Let }u=x+2 \implies du = dx} \\ & = \int_2^6 (u-4)\sqrt{u^2 -3} \ du & \small \blue{\text{Let }u = \sqrt 3 \cosh t \implies du = \sqrt 3 \sinh t \ dt} \\ & = \int_{\cosh^{-1} \frac 2{\sqrt 3}}^{\cosh^{-1} 2\sqrt 3} 3(\sqrt 3 \cosh t - 4) \sinh^2 t \ dt \\ & = 3\sqrt 3 \int_{\cosh^{-1} \frac 2{\sqrt 3}}^{\cosh^{-1} 2\sqrt 3} \cosh t \sinh^2 t \ dt - 12 \int_{\cosh^{-1} \frac 2{\sqrt 3}}^{\cosh^{-1} 2\sqrt 3} \sinh^2 t \ dt \\ & = 3\sqrt 3 \int_{\frac 1{\sqrt 3}}^{\sqrt {11}} \sinh^2 t \ d \sinh t - 6 \int_{\cosh^{-1} \frac 2{\sqrt 3}}^{\cosh^{-1} 2\sqrt 3} (\cosh 2t -1) \ dt \\ & = \sqrt 3 \sinh^3 t \bigg|_{\frac 1{\sqrt 3}}^{\sqrt {11}} - 6 \left[\frac {\sinh 2t}2 - t \right]_{\cosh^{-1} \frac 2{\sqrt 3}}^{\cosh^{-1} 2\sqrt 3} \\ & = 11\sqrt{33} - \frac 13 - 12\sqrt{33} + 4 + 6 \cosh^{-1} 2\sqrt 3 - 6 \cosh^{-1} \frac 2{\sqrt 3} \\ & = \frac {11}3 - \sqrt{33} + 6\ln(6+\sqrt{33}) - 6 \ln 3 \\ & \approx 6.110772568 \end{aligned}$

Therefore $\lfloor 10000 A \rfloor = \boxed{61107}$ .