Chọn phương án B.

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

Theo đó \(a=2,\,b=3,\,c=2\).
Suy ra \(a+b+c=7\).