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Câu hỏi: Giả sử $\int\limits_{1}^{2}{\left( 2x-1 \right)\ln x\text{d}x}=a\ln 2+b$, $\left( a;b\in \mathbb{Q} \right)$. Tính $a+b$.
A. $\dfrac{5}{2}$.
B. $2$.
C. $1$.
D. $\dfrac{3}{2}$.
A. $\dfrac{5}{2}$.
B. $2$.
C. $1$.
D. $\dfrac{3}{2}$.
Đặt
$\left\{ \begin{aligned}
& u=\ln x \\
& dv=\left( 2x-1 \right)dx \\
\end{aligned} \right. $ $ \Rightarrow \left\{ \begin{aligned}
& du=\dfrac{1}{x}dx \\
& v={{x}^{2}}-x \\
\end{aligned} \right.$
$\int\limits_{1}^{2}{\left( 2x-1 \right)\ln x}\text{d}x=$ $=\left. \left[ \left( {{x}^{2}}-x \right)\ln x \right] \right|_{1}^{2}-\int\limits_{1}^{2}{\dfrac{{{x}^{2}}-x}{x}}\text{d}x$ $=2\ln 2-\left. \left( \dfrac{{{x}^{2}}}{2}-x \right) \right|_{1}^{2}$ $=2\ln 2-\dfrac{1}{2}$ nên $a=2$, $b=-\dfrac{1}{2}$.
Vậy $a+b$ $=\dfrac{3}{2}$.
$\left\{ \begin{aligned}
& u=\ln x \\
& dv=\left( 2x-1 \right)dx \\
\end{aligned} \right. $ $ \Rightarrow \left\{ \begin{aligned}
& du=\dfrac{1}{x}dx \\
& v={{x}^{2}}-x \\
\end{aligned} \right.$
$\int\limits_{1}^{2}{\left( 2x-1 \right)\ln x}\text{d}x=$ $=\left. \left[ \left( {{x}^{2}}-x \right)\ln x \right] \right|_{1}^{2}-\int\limits_{1}^{2}{\dfrac{{{x}^{2}}-x}{x}}\text{d}x$ $=2\ln 2-\left. \left( \dfrac{{{x}^{2}}}{2}-x \right) \right|_{1}^{2}$ $=2\ln 2-\dfrac{1}{2}$ nên $a=2$, $b=-\dfrac{1}{2}$.
Vậy $a+b$ $=\dfrac{3}{2}$.
Đáp án D.