Chọn phương án A.

Đặt $t=\sqrt{x^2+4}\Leftrightarrow t^2=x^2+4\Rightarrow t\mathrm{\,d}t=x\mathrm{\,d}x$.

• $x=0\Rightarrow t=2$
• $x=1\Rightarrow t=\sqrt{5}$

$\Rightarrow\displaystyle\int\limits_0^1x\sqrt{x^2+4}\mathrm{\,d}x=\displaystyle\int\limits_2^{\sqrt{5}}t^2\mathrm{\,d}t=\dfrac{t^3}{3}\bigg|_2^{\sqrt{5}}=\dfrac{1}{3}\left(\sqrt{5^3}-8\right)$.

Theo đó $\begin{cases}
a=3\\ b=5\\ c=8
\end{cases}\Rightarrow Q=abc=120$.