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Câu hỏi: Biết rằng $\int\limits_{2}^{4}{\dfrac{{{x}^{3}}+2}{{{x}^{2}}+x}dx}=a+b\ln 2+c\ln 3+d\ln 5,$ với $a,b,c,d\in \mathbb{Z}.$ Tính giá trị của biểu thức $S=a+b+c+d.$
A. $S=6.$
B. $S=8.$
C. $S=10.$
D. $S=4.$
A. $S=6.$
B. $S=8.$
C. $S=10.$
D. $S=4.$
Phân tích $\dfrac{{{x}^{3}}+2}{{{x}^{2}}+x}=\dfrac{x\left( {{x}^{2}}+x \right)-\left( {{x}^{2}}+x \right)+x+2}{{{x}^{2}}+x}=x-1+\dfrac{x+2}{x\left( x+1 \right)}=x-1+\dfrac{m}{x}+\dfrac{n}{x+1}$
$\Rightarrow x+2=m\left( x+1 \right)+nx,$ cho $\left\{ \begin{aligned}
& x=0\Rightarrow m=2 \\
& x=-1\Rightarrow n=-1 \\
\end{aligned} \right.\Rightarrow \int\limits_{2}^{4}{\dfrac{{{x}^{3}}+2}{{{x}^{2}}+x}dx}=\int\limits_{2}^{4}{\left( x-1+\dfrac{2}{x}-\dfrac{1}{x+1} \right)dx}$
$\Rightarrow I=\left. \left( \dfrac{{{x}^{2}}}{2}-x+2\ln \left| x \right|-\ln \left| x+1 \right| \right) \right|_{2}^{4}=\left( 4+2\ln 4-\ln 5 \right)-\left( 2\ln 2-\ln 3 \right)$
$=4+2\ln 2+\ln 3-\ln 5\Rightarrow a=4,b=2,c=1,d=-1\Rightarrow S=6.$
$\Rightarrow x+2=m\left( x+1 \right)+nx,$ cho $\left\{ \begin{aligned}
& x=0\Rightarrow m=2 \\
& x=-1\Rightarrow n=-1 \\
\end{aligned} \right.\Rightarrow \int\limits_{2}^{4}{\dfrac{{{x}^{3}}+2}{{{x}^{2}}+x}dx}=\int\limits_{2}^{4}{\left( x-1+\dfrac{2}{x}-\dfrac{1}{x+1} \right)dx}$
$\Rightarrow I=\left. \left( \dfrac{{{x}^{2}}}{2}-x+2\ln \left| x \right|-\ln \left| x+1 \right| \right) \right|_{2}^{4}=\left( 4+2\ln 4-\ln 5 \right)-\left( 2\ln 2-\ln 3 \right)$
$=4+2\ln 2+\ln 3-\ln 5\Rightarrow a=4,b=2,c=1,d=-1\Rightarrow S=6.$
Đáp án A.