Chọn phương án B.

\(\begin{aligned}\dfrac{1}{(x+1)(2x+1)}&=\dfrac{2(x+1)-(2x+1)}{(x+1)(2x+1)}\\ &=\dfrac{2}{2x+1}-\dfrac{1}{x+1}.\end{aligned}\)

Khi đó ta có $$\begin{aligned}
\displaystyle\int\limits_{1}^{2}\dfrac{\mathrm{d}x}{(x+1)(2x+1)}&=\displaystyle\int\limits_{1}^{2}\left(\dfrac{2}{2x+1}-\dfrac{1}{x+1}\right)\mathrm{d}x\\
&=\left(\ln|2x+1|-\ln|x+1|\right)\bigg|_1^2\\
&=\left(\ln5-\ln3\right)-\left(\ln3-\ln2\right)\\
&=\ln2-2\ln3+\ln5.
\end{aligned}$$
Vậy \(a=1\), \(b=-2\), \(c=1\).

Do đó \(a+b+c=1-2+1=0\).