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Câu hỏi: Biết rằng $\int\limits_{2}^{3}{\dfrac{x+1}{x\left( x-2 \right)+1}dx}=a+b\ln 2,$ với $a,b\in \mathbb{Z}.$ Tính $S=a+2b.$
A. $S=1.$
B. $S=4.$
C. $S=3.$
D. $S=5.$
A. $S=1.$
B. $S=4.$
C. $S=3.$
D. $S=5.$
Ta có $\int\limits_{2}^{3}{\dfrac{x+1}{x\left( x-2 \right)+1}dx}=\int\limits_{2}^{3}{\dfrac{x+1}{{{x}^{2}}-2x+1}dx}=\int\limits_{2}^{3}{\dfrac{x+1}{{{\left( x-1 \right)}^{2}}}dx}=\int\limits_{2}^{3}{\dfrac{x-1+2}{{{\left( x-1 \right)}^{2}}}dx}$
$\begin{aligned}
& =\int\limits_{2}^{3}{\left[ \dfrac{1}{x-1}+\dfrac{2}{{{\left( x-1 \right)}^{2}}} \right]dx=\left( \ln \left| x-1 \right|-\dfrac{2}{x-1} \right)\left| \begin{aligned}
& ^{3} \\
& _{2} \\
\end{aligned} \right.} \\
& =\left( \ln 2-1 \right)-\left( -2 \right)=1+\ln 2\Rightarrow \left\{ \begin{aligned}
& a=1 \\
& b=1 \\
\end{aligned} \right.\Rightarrow S=3. \\
\end{aligned}$
$\begin{aligned}
& =\int\limits_{2}^{3}{\left[ \dfrac{1}{x-1}+\dfrac{2}{{{\left( x-1 \right)}^{2}}} \right]dx=\left( \ln \left| x-1 \right|-\dfrac{2}{x-1} \right)\left| \begin{aligned}
& ^{3} \\
& _{2} \\
\end{aligned} \right.} \\
& =\left( \ln 2-1 \right)-\left( -2 \right)=1+\ln 2\Rightarrow \left\{ \begin{aligned}
& a=1 \\
& b=1 \\
\end{aligned} \right.\Rightarrow S=3. \\
\end{aligned}$
Đáp án C.