Chọn phương án B.

\(\begin{eqnarray*}
&&\displaystyle\int\limits_1^2\left(x^2+\dfrac{x}{x+1}\right)\mathrm{\,d}x\\
&=&\displaystyle\int\limits_1^2\left(x^2+1-\dfrac{1}{x+1}\right)\mathrm{\,d}x\\
&=&\left.\left(\dfrac{x^3}{3}+x-\ln\left|x+1\right|\right)\right|_1^2\\
&=&\left(\dfrac{8}{3}+2-\ln 3\right)-\left(\frac{1}{3}+1-\ln 2\right)\\
&=&\dfrac{10}{3}+\ln\dfrac{2}{3}.
\end{eqnarray*}\)

Theo đó \(a=2,\,b=3\Rightarrow P=a+b=5\).