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Câu hỏi: Biết rằng $\int\limits_{1}^{2}{\dfrac{x-1}{x\left( x+1 \right)\left( x+2 \right)}dx=a\ln 2+b\ln 3}$ với a, $b\in \mathbb{Q}$. Tính $P=a+b$.
A. $P=3$.
B. $P=5$.
C. $P=-2$.
D. $P=-1$.
A. $P=3$.
B. $P=5$.
C. $P=-2$.
D. $P=-1$.
Ta có $\dfrac{x-1}{x\left( x+1 \right)\left( x+2 \right)}=\dfrac{a}{x}+\dfrac{b}{x+1}+\dfrac{c}{x+2}\Rightarrow x-1=a\left( x+1 \right)\left( x+2 \right)+bx\left( x+2 \right)+cx\left( x+1 \right)$
$\Rightarrow a=\dfrac{x-1}{\left( x+1 \right)\left( x+2 \right)}\left| \begin{aligned}
& \\
& _{x=0} \\
\end{aligned} \right.=-\dfrac{1}{2} $; $ b=\dfrac{x-1}{x\left( x+2 \right)}\left| \begin{aligned}
& \\
& _{x=-1} \\
\end{aligned} \right.=2 $; $ c=\dfrac{x-1}{x\left( x+1 \right)}\left| \begin{aligned}
& \\
& _{x=-2} \\
\end{aligned} \right.=-\dfrac{3}{2}$
$\Rightarrow \int\limits_{1}^{2}{\dfrac{x-1}{x\left( x+1 \right)\left( x+2 \right)}dx}=-\dfrac{1}{2}\int\limits_{1}^{2}{\dfrac{1}{x}dx}+2\int\limits_{1}^{2}{\dfrac{1}{x+1}dx}-\dfrac{3}{2}\int\limits_{1}^{2}{\dfrac{1}{x+2}dx}$
$=-\dfrac{1}{2}\ln \left| x \right|\left| \begin{aligned}
& ^{2} \\
& _{1} \\
\end{aligned} \right.+2\ln \left| x+1 \right|\left| \begin{aligned}
& ^{2} \\
& _{1} \\
\end{aligned} \right.-\dfrac{3}{2}\ln \left| x+2 \right|\left| \begin{aligned}
& ^{2} \\
& _{1} \\
\end{aligned} \right.$
$=-\dfrac{1}{2}\ln 2+2\ln 3-2\ln 2-\dfrac{3}{2}\ln 4+\dfrac{3}{2}\ln 3=-\dfrac{11}{2}\ln 2+\dfrac{7}{2}\ln 3$
$\Rightarrow a=\dfrac{x-1}{\left( x+1 \right)\left( x+2 \right)}\left| \begin{aligned}
& \\
& _{x=0} \\
\end{aligned} \right.=-\dfrac{1}{2} $; $ b=\dfrac{x-1}{x\left( x+2 \right)}\left| \begin{aligned}
& \\
& _{x=-1} \\
\end{aligned} \right.=2 $; $ c=\dfrac{x-1}{x\left( x+1 \right)}\left| \begin{aligned}
& \\
& _{x=-2} \\
\end{aligned} \right.=-\dfrac{3}{2}$
$\Rightarrow \int\limits_{1}^{2}{\dfrac{x-1}{x\left( x+1 \right)\left( x+2 \right)}dx}=-\dfrac{1}{2}\int\limits_{1}^{2}{\dfrac{1}{x}dx}+2\int\limits_{1}^{2}{\dfrac{1}{x+1}dx}-\dfrac{3}{2}\int\limits_{1}^{2}{\dfrac{1}{x+2}dx}$
$=-\dfrac{1}{2}\ln \left| x \right|\left| \begin{aligned}
& ^{2} \\
& _{1} \\
\end{aligned} \right.+2\ln \left| x+1 \right|\left| \begin{aligned}
& ^{2} \\
& _{1} \\
\end{aligned} \right.-\dfrac{3}{2}\ln \left| x+2 \right|\left| \begin{aligned}
& ^{2} \\
& _{1} \\
\end{aligned} \right.$
$=-\dfrac{1}{2}\ln 2+2\ln 3-2\ln 2-\dfrac{3}{2}\ln 4+\dfrac{3}{2}\ln 3=-\dfrac{11}{2}\ln 2+\dfrac{7}{2}\ln 3$
Đáp án C.