Chọn phương án A.

$\begin{aligned}
\displaystyle\int\limits_0^1\dfrac{\mathrm{d}x}{\sqrt{x+1}+\sqrt{x}}&=\displaystyle\int\limits_0^1\left(\sqrt{x+1}-\sqrt{x}\right)\mathrm{d}x\\
&=\displaystyle\int\limits_0^1\left((x+1)^{\tfrac{1}{2}}-x^{\tfrac{1}{2}}\right)\mathrm{d}x\\
&=\dfrac{2}{3}\left(\sqrt{(x+1)^3}-\sqrt{x^3}\right)\bigg|_0^1\\
&=\dfrac{2}{3}\left(\sqrt{8}-2\right).
\end{aligned}$

Theo đó $\begin{cases}
a=8\\ b=2
\end{cases}\Rightarrow T=a+b=10$.