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Câu hỏi: Cho Tích phân $I=\int\limits_{1}^{2}{\left( 2x-1 \right)\ln x\text{d}x}$ bằng
A. $I=2\ln 2+\dfrac{1}{2}$
B. $I=\dfrac{1}{2}.$
C. $I=2\ln 2.$
D. $I=2\ln 2-\dfrac{1}{2}.$
A. $I=2\ln 2+\dfrac{1}{2}$
B. $I=\dfrac{1}{2}.$
C. $I=2\ln 2.$
D. $I=2\ln 2-\dfrac{1}{2}.$
Đặt $\left\{ \begin{aligned}
& u=\ln x \\
& dv=\left( 2x-1 \right)\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{1}{x} \\
& v={{x}^{2}}-x \\
\end{aligned} \right.$
Do đó $I=\left. \left( {{x}^{2}}-x \right)\ln x \right|_{1}^{2}-\int\limits_{1}^{2}{\left( x-1 \right)dx}$ $=2\ln 2-\left. \left( \dfrac{{{x}^{2}}}{2}-x \right) \right|_{1}^{2}=2\ln 2-\dfrac{1}{2}$.
& u=\ln x \\
& dv=\left( 2x-1 \right)\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{1}{x} \\
& v={{x}^{2}}-x \\
\end{aligned} \right.$
Do đó $I=\left. \left( {{x}^{2}}-x \right)\ln x \right|_{1}^{2}-\int\limits_{1}^{2}{\left( x-1 \right)dx}$ $=2\ln 2-\left. \left( \dfrac{{{x}^{2}}}{2}-x \right) \right|_{1}^{2}=2\ln 2-\dfrac{1}{2}$.
Đáp án D.