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Câu hỏi: Cho $\int\limits_{1}^{e}{\left( 2+x\ln x \right)dx=a{{e}^{2}}+be+c}$ với $a,b,c$ là các số hữu tỉ. Mệnh đề nào sau đây đúng?
A. $a+b=-c$
B. $a+b=c$
C. $a-b=c$
D. $a-b=-c$
A. $a+b=-c$
B. $a+b=c$
C. $a-b=c$
D. $a-b=-c$
Ta có $\int\limits_{1}^{e}{\left( 2+x\ln x \right)dx=\int\limits_{1}^{e}{2dx+\int\limits_{1}^{e}{x\ln xdx=\left. 2x \right|_{1}^{e}+I=2e-2+I}}}$ với $I=\int\limits_{1}^{e}{x\ln xdx}$
Đặt $\left\{ \begin{aligned}
& u=\ln x \\
& dv=xdx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{1}{x}dx \\
& v=\dfrac{{{x}^{2}}}{2} \\
\end{aligned} \right.\Rightarrow I=\left. \dfrac{{{x}^{2}}}{2}\ln x \right|_{1}^{e}-\int\limits_{1}^{e}{\dfrac{x}{2}dx}=\left. \dfrac{{{x}^{2}}}{2}\ln x \right|_{1}^{e}-\left. \dfrac{{{x}^{2}}}{4} \right|_{1}^{e}$
$=\dfrac{{{e}^{2}}}{2}-\dfrac{1}{4}\left( {{e}^{2}}-1 \right)=\dfrac{{{e}^{2}}+1}{4}$
$\Rightarrow \int\limits_{1}^{e}{\left( 2+x\ln x \right)dx=2e-2+\dfrac{{{e}^{2}}+1}{4}=\dfrac{1}{4}{{e}^{2}}+2e-\dfrac{7}{4}\Rightarrow \left\{ \begin{aligned}
& a=\dfrac{1}{4} \\
& b=2 \\
& c=-\dfrac{7}{4} \\
\end{aligned} \right.\Rightarrow a-b=c}$
Đặt $\left\{ \begin{aligned}
& u=\ln x \\
& dv=xdx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{1}{x}dx \\
& v=\dfrac{{{x}^{2}}}{2} \\
\end{aligned} \right.\Rightarrow I=\left. \dfrac{{{x}^{2}}}{2}\ln x \right|_{1}^{e}-\int\limits_{1}^{e}{\dfrac{x}{2}dx}=\left. \dfrac{{{x}^{2}}}{2}\ln x \right|_{1}^{e}-\left. \dfrac{{{x}^{2}}}{4} \right|_{1}^{e}$
$=\dfrac{{{e}^{2}}}{2}-\dfrac{1}{4}\left( {{e}^{2}}-1 \right)=\dfrac{{{e}^{2}}+1}{4}$
$\Rightarrow \int\limits_{1}^{e}{\left( 2+x\ln x \right)dx=2e-2+\dfrac{{{e}^{2}}+1}{4}=\dfrac{1}{4}{{e}^{2}}+2e-\dfrac{7}{4}\Rightarrow \left\{ \begin{aligned}
& a=\dfrac{1}{4} \\
& b=2 \\
& c=-\dfrac{7}{4} \\
\end{aligned} \right.\Rightarrow a-b=c}$
Đáp án C.