Chọn phương án D.

\(\begin{eqnarray*}
&&\displaystyle\int\limits_1^2\frac{x}{(x+1)^2}\mathrm{\,d}x\\
&=&\displaystyle\int\limits_1^2\frac{(x+1)-1}{(x+1)^2}\mathrm{\,d}x\\
&=&\displaystyle\int\limits_1^2\left(\dfrac{1}{x+1}-\dfrac{1}{(x+1)^2}\right)\mathrm{\,d}x\\
&=&\displaystyle\int\limits_1^2\dfrac{1}{x+1}\mathrm{\,d}x-\displaystyle\int\limits_1^2\dfrac{1}{(x+1)^2}\mathrm{\,d}x\\
&=&\ln\left|x+1\right|\bigg|_1^2+\dfrac{1}{x+1}\bigg|_1^2\\
&=&\ln 3-\ln 2+\dfrac{1}{3}-\dfrac{1}{2}\\
&=&-\dfrac{1}{6}-\ln 2+\ln 3.\end{eqnarray*}\)

Theo đó \(a=-\dfrac{1}{6},\,b=-1,\,c=1\)
Suy ra \(6a+b+c=6\cdot\dfrac{-1}{6}+(-1)+1=-1\).