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Câu hỏi: [ Mức độ 2] Giả sử $\int\limits_{0}^{2}{\dfrac{x-1}{{{x}^{2}}+4x+3}}\text{d}x=a\ln 5+b\ln 3;\ a,b\in \mathbb{Q}$. Tính $P=a.b$.
A. $P=-5$.
B. $P=-4$.
C. $P=8$.
D. $P=-6$.
A. $P=-5$.
B. $P=-4$.
C. $P=8$.
D. $P=-6$.
Ta có $\dfrac{x-1}{{{x}^{2}}+4x+3}=\dfrac{x-1}{(x+3)(x+1)}=\dfrac{A}{x+3}+\dfrac{B}{x+1}$ $=\dfrac{(A+B)x+A+3B}{(x+3)(x+1)}$.
Đồng nhất thức ta có $\left\{ \begin{aligned}
& A+B=1 \\
& A+3B=-1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& B=-1 \\
& A=2 \\
\end{aligned} \right.$.
Suy ra $\dfrac{x-1}{{{x}^{2}}+4x+3}=\dfrac{2}{x+3}-\dfrac{1}{x+1}$.
Suy ra $\int\limits_{0}^{2}{\dfrac{x-1}{{{x}^{2}}+4x+3}}\text{d}x=\int\limits_{0}^{2}{\left( \dfrac{2}{x+3}-\dfrac{1}{x+1} \right)}\text{d}x$
$=2\int\limits_{0}^{2}{\dfrac{\text{d}x}{x+3}}-\int\limits_{0}^{2}{\dfrac{\text{d}x}{x+1}}=\left. 2\ln \left| x+3 \right| \right|_{0}^{2}-\left. \ln \left| x+1 \right| \right|_{0}^{2}$
$=2\ln 5-2\ln 3-\ln 3=2\ln 5-3\ln 3$.
Suy ra $a=2;\ b=-3$. Vậy $P=a.b=-6$.
Đồng nhất thức ta có $\left\{ \begin{aligned}
& A+B=1 \\
& A+3B=-1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& B=-1 \\
& A=2 \\
\end{aligned} \right.$.
Suy ra $\dfrac{x-1}{{{x}^{2}}+4x+3}=\dfrac{2}{x+3}-\dfrac{1}{x+1}$.
Suy ra $\int\limits_{0}^{2}{\dfrac{x-1}{{{x}^{2}}+4x+3}}\text{d}x=\int\limits_{0}^{2}{\left( \dfrac{2}{x+3}-\dfrac{1}{x+1} \right)}\text{d}x$
$=2\int\limits_{0}^{2}{\dfrac{\text{d}x}{x+3}}-\int\limits_{0}^{2}{\dfrac{\text{d}x}{x+1}}=\left. 2\ln \left| x+3 \right| \right|_{0}^{2}-\left. \ln \left| x+1 \right| \right|_{0}^{2}$
$=2\ln 5-2\ln 3-\ln 3=2\ln 5-3\ln 3$.
Suy ra $a=2;\ b=-3$. Vậy $P=a.b=-6$.
Đáp án D.