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Câu hỏi: Cho $\int\limits_{1}^{e}{\left( 1+x\ln x \right)dx}=a{{e}^{2}}+be+c$ với $a,b,c$ là các số hữu tỷ. Mệnh đề nào dưới 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 $\int\limits_{1}^{e}{\left( 1+x\ln x \right)dx}=\int\limits_{1}^{e}{1.dx}+\int\limits_{1}^{e}{x\ln xdx}=e-1+\int\limits_{1}^{e}{x\ln xdx}.$
Đặt $\left\{ \begin{aligned}
& u=\ln x\Rightarrow du=\dfrac{1}{x}dx \\
& dv=x.dx\Rightarrow v=\dfrac{{{x}^{2}}}{2} \\
\end{aligned} \right.. $ Khi đó $ \int\limits_{1}^{e}{x\ln xdx}=\left. \dfrac{{{x}^{2}}}{2}\ln x \right|_{1}^{e}-\dfrac{1}{2}\int\limits_{1}^{e}{xdx}=\left. \dfrac{{{e}^{2}}}{2}-\dfrac{1}{4}{{x}^{2}} \right|_{1}^{e}$
$=\dfrac{{{e}^{2}}}{2}-\dfrac{{{e}^{2}}}{4}+\dfrac{1}{4}=\dfrac{{{e}^{2}}}{4}+\dfrac{1}{4}.$ Suy ra $\int\limits_{1}^{e}{\left( 1+x\ln x \right)dx}=e-1+\dfrac{{{e}^{2}}}{4}+\dfrac{1}{4}=\dfrac{{{e}^{2}}}{4}+e-\dfrac{3}{4}$ nên $a=\dfrac{1}{4},b=1,c=-\dfrac{3}{4}.$ Vậy $a-b=c.$
Đặt $\left\{ \begin{aligned}
& u=\ln x\Rightarrow du=\dfrac{1}{x}dx \\
& dv=x.dx\Rightarrow v=\dfrac{{{x}^{2}}}{2} \\
\end{aligned} \right.. $ Khi đó $ \int\limits_{1}^{e}{x\ln xdx}=\left. \dfrac{{{x}^{2}}}{2}\ln x \right|_{1}^{e}-\dfrac{1}{2}\int\limits_{1}^{e}{xdx}=\left. \dfrac{{{e}^{2}}}{2}-\dfrac{1}{4}{{x}^{2}} \right|_{1}^{e}$
$=\dfrac{{{e}^{2}}}{2}-\dfrac{{{e}^{2}}}{4}+\dfrac{1}{4}=\dfrac{{{e}^{2}}}{4}+\dfrac{1}{4}.$ Suy ra $\int\limits_{1}^{e}{\left( 1+x\ln x \right)dx}=e-1+\dfrac{{{e}^{2}}}{4}+\dfrac{1}{4}=\dfrac{{{e}^{2}}}{4}+e-\dfrac{3}{4}$ nên $a=\dfrac{1}{4},b=1,c=-\dfrac{3}{4}.$ Vậy $a-b=c.$
Đáp án C.