Chọn: Đáp án D

$I=\int{\frac{dx}{2{{x}^{2}}-x-1}=}\int{\frac{dx}{\left( x-1 \right)\left( 2x+1 \right)}=}\int{\frac{\left( 2x+1 \right)-2\left( x-1 \right)}{\left( x-1 \right)\left( 2x+1 \right)}dx}$

$I=\int{\frac{1}{3}\left( \frac{1}{x-1}-\frac{2}{2x+1} \right)dx=}\frac{1}{3}\ln \left| x-1 \right|-\frac{2}{3}\ln \left| x-1 \right|+C$

Khi đó $a=\frac{1}{3},b=-\frac{2}{3},c=1\Rightarrow 2a+b=0$