Chọn phương án B.

Đặt $A=\displaystyle\int\limits_3^4\dfrac{\mathrm{\,d}x}{x^2+x}$ ta có $$A=a\ln2+b\ln3+c\ln5=\ln2^a3^b5^c\Leftrightarrow\mathrm{e}^A=2^a3^b5^c.$$

1. Gán giá trị $\displaystyle\int\limits_3^4\dfrac{\mathrm{\,d}x}{x^2+x}$ vào biến nhớ A.
   
2. Khi đó $\mathrm{e}^A=\dfrac{16}{15}$.
   

Vậy $2^a3^b5^c=\dfrac{16}{15}=16\cdot\dfrac{1}{3}\cdot\dfrac{1}{5}=2^4\cdot3^{-1}\cdot5^{-1}$.

Do đó $a=4$, $b=c=-1$. Suy ra $a+b+c=2$.

Chọn phương án B.

\(\begin{eqnarray*}
&I&=\displaystyle\int\limits_3^4\dfrac{\mathrm{\,d}x}{x^2+x}=\displaystyle\int\limits_3^4\dfrac{\mathrm{\,d}x}{x(x+1)}\\
&&=\displaystyle\int\limits_3^4\left(\dfrac{1}{x}-\dfrac{1}{x+1}\right)\mathrm{\,d}x\\
&&=\displaystyle\int\limits_3^4\dfrac{\mathrm{\,d}x}{x}
-\displaystyle\int\limits_3^4\dfrac{\mathrm{\,d}x}{x+1}\\
&&=\ln|x|\bigg|_3^4-\ln|x+1|\bigg|_3^4\\
&&=\left(\ln4-\ln3\right)-\left(\ln5+\ln4\right)\\
&&=4\ln 2-\ln 3-\ln 5.\end{eqnarray*}\)

Theo đó \(a=4,\,b=c=-1\Rightarrow S=2\).