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Câu hỏi: Cho biết $\int\limits_{1}^{5}{\dfrac{3dx}{{{x}^{2}}+3x}}=a\ln 5+b\ln 2\left( a,b\in \mathbb{Z} \right)$. Mệnh đề nào sau đây đúng
A. $2a-b=0$.
B. $a-b=0$.
C. $a+2b=0$.
D. $a+b=0$.
A. $2a-b=0$.
B. $a-b=0$.
C. $a+2b=0$.
D. $a+b=0$.
Xét $\dfrac{3}{{{x}^{2}}+3x}=\dfrac{3}{x\left( x+3 \right)}=\dfrac{A}{x}+\dfrac{B}{x+3}=\dfrac{A\left( x+3 \right)+Bx}{x\left( x+3 \right)}$
$\Leftrightarrow 3=Ax+Bx+3.A\Leftrightarrow 0x+3=\left( A+B \right)x+3.A\Rightarrow \left\{ \begin{aligned}
& A+B=0 \\
& 3A=3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& A=1 \\
& B=-1 \\
\end{aligned} \right.$
$\Rightarrow \int\limits_{1}^{5}{\dfrac{3dx}{{{x}^{2}}+3x}}=\int\limits_{1}^{5}{\left( \dfrac{1}{x}-\dfrac{1}{x+3} \right)dx}=\left. \left( \ln x-\ln \left( x+3 \right) \right) \right|_{1}^{5}=\ln 5-\ln 8-\ln 1+\ln 4$
$=\ln 5-3\ln 2+2\ln 2=\ln 5-\ln 2\Rightarrow \left\{ \begin{aligned}
& a=1 \\
& b=-1 \\
\end{aligned} \right.\Rightarrow a+b=0$.
$\Leftrightarrow 3=Ax+Bx+3.A\Leftrightarrow 0x+3=\left( A+B \right)x+3.A\Rightarrow \left\{ \begin{aligned}
& A+B=0 \\
& 3A=3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& A=1 \\
& B=-1 \\
\end{aligned} \right.$
$\Rightarrow \int\limits_{1}^{5}{\dfrac{3dx}{{{x}^{2}}+3x}}=\int\limits_{1}^{5}{\left( \dfrac{1}{x}-\dfrac{1}{x+3} \right)dx}=\left. \left( \ln x-\ln \left( x+3 \right) \right) \right|_{1}^{5}=\ln 5-\ln 8-\ln 1+\ln 4$
$=\ln 5-3\ln 2+2\ln 2=\ln 5-\ln 2\Rightarrow \left\{ \begin{aligned}
& a=1 \\
& b=-1 \\
\end{aligned} \right.\Rightarrow a+b=0$.
Đáp án D.