Chọn phương án B.

Đặt \(u=x^2+1\) ta có

• \(\mathrm{d}u=2x\mathrm{\,d}x\Leftrightarrow x\mathrm{\,d}x=\dfrac{1}{2}\mathrm{\,d}u\)
• \(x=2\Rightarrow u=5\)
• \(x=7\Rightarrow u=50\)

Suy ra $$\begin{aligned}\displaystyle\int\limits_2^7\dfrac{x\mathrm{\,d}x}{x^2+1}&=\displaystyle\int_{5}^{50}\dfrac{1}{2u}\mathrm{\,d}u\\
&=\dfrac{\ln|u|}{2}\bigg|_5^{50}\\
&=\dfrac{1}{2}\ln2+\dfrac{1}{2}\ln5.\end{aligned}$$
Do đó \(a=\dfrac{1}{2}\) và \(b=-\dfrac{1}{2}\).

Vậy \(2a+b=\dfrac{1}{2}\).