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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: $\int\limits_{1}^{e}{\left( 2+x\ln x \right)dx=\int\limits_{1}^{e}{2dx+\int\limits_{1}^{e}{x\ln xdx=\left. 2x \right|_{1}^{e}+I=2e-2+I}}}$.
Tính $I$ :
Đặt $\left\{ \begin{aligned}
& u=\ln x\Rightarrow du=\dfrac{1}{x}.dx \\
& dv=xdx\Rightarrow v=\dfrac{{{x}^{2}}}{2} \\
\end{aligned} \right.$
$\Rightarrow I=\left. \dfrac{{{x}^{2}}}{2}.\ln x \right|_{1}^{e}-\int\limits_{1}^{e}{\dfrac{{{x}^{2}}}{2}.\dfrac{1}{x}dx=\dfrac{{{e}^{2}}}{2}-\int\limits_{1}^{e}{\dfrac{x}{2}dx=\dfrac{{{e}^{2}}}{2}-\left. \left( \dfrac{{{x}^{2}}}{4} \right) \right|}}_{1}^{e}=\dfrac{{{e}^{2}}}{2}-\dfrac{{{e}^{2}}}{4}+\dfrac{1}{4}=\dfrac{{{e}^{2}}}{4}+\dfrac{1}{4}$
Vậy $\int\limits_{1}^{e}{\left( 2+x\ln x \right)dx=2e-2+\dfrac{{{e}^{2}}}{4}+\dfrac{1}{4}=\dfrac{{{e}^{2}}}{4}+2e-\dfrac{7}{4}}$
$\Rightarrow a=\dfrac{1}{4};b=2;c=-\dfrac{7}{4}$.
Tính $I$ :
Đặt $\left\{ \begin{aligned}
& u=\ln x\Rightarrow du=\dfrac{1}{x}.dx \\
& dv=xdx\Rightarrow v=\dfrac{{{x}^{2}}}{2} \\
\end{aligned} \right.$
$\Rightarrow I=\left. \dfrac{{{x}^{2}}}{2}.\ln x \right|_{1}^{e}-\int\limits_{1}^{e}{\dfrac{{{x}^{2}}}{2}.\dfrac{1}{x}dx=\dfrac{{{e}^{2}}}{2}-\int\limits_{1}^{e}{\dfrac{x}{2}dx=\dfrac{{{e}^{2}}}{2}-\left. \left( \dfrac{{{x}^{2}}}{4} \right) \right|}}_{1}^{e}=\dfrac{{{e}^{2}}}{2}-\dfrac{{{e}^{2}}}{4}+\dfrac{1}{4}=\dfrac{{{e}^{2}}}{4}+\dfrac{1}{4}$
Vậy $\int\limits_{1}^{e}{\left( 2+x\ln x \right)dx=2e-2+\dfrac{{{e}^{2}}}{4}+\dfrac{1}{4}=\dfrac{{{e}^{2}}}{4}+2e-\dfrac{7}{4}}$
$\Rightarrow a=\dfrac{1}{4};b=2;c=-\dfrac{7}{4}$.
Đáp án A.