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Câu hỏi: Biết tích phân $I=\int\limits_{1}^{e}{x\ln xdx=\dfrac{a{{e}^{2}}+b}{c},\left( a,b,c\in Z \right)}$. Giá trị của a+b+c là
A. 7.
B. 4.
C. 8.
D. 6.
A. 7.
B. 4.
C. 8.
D. 6.
$I=\int\limits_{1}^{e}{x\ln \text{xdx}.}$ Đặt $\left\{ \begin{aligned}
& u=\ln \text{x} \\
& dv=x\text{dx} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{1}{x}d\text{x} \\
& v=\dfrac{{{x}^{2}}}{2} \\
\end{aligned} \right.$
$\Rightarrow I=\dfrac{{{x}^{2}}}{2}\ln \text{x}\mathop{|}_{1}^{e}-\int\limits_{1}^{e}{\dfrac{1}{x}.\dfrac{{{x}^{2}}}{2}d\text{x}=\dfrac{{{e}^{2}}}{2}-\dfrac{1}{2}\int\limits_{1}^{e}{x\text{dx}=\dfrac{{{e}^{2}}}{2}-\dfrac{{{x}^{2}}}{4}\mathop{|}_{0}^{e}}}=\dfrac{{{e}^{2}}}{2}-\dfrac{{{e}^{2}}}{4}+\dfrac{1}{4}=\dfrac{{{e}^{2}}+1}{4}.$
& u=\ln \text{x} \\
& dv=x\text{dx} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{1}{x}d\text{x} \\
& v=\dfrac{{{x}^{2}}}{2} \\
\end{aligned} \right.$
$\Rightarrow I=\dfrac{{{x}^{2}}}{2}\ln \text{x}\mathop{|}_{1}^{e}-\int\limits_{1}^{e}{\dfrac{1}{x}.\dfrac{{{x}^{2}}}{2}d\text{x}=\dfrac{{{e}^{2}}}{2}-\dfrac{1}{2}\int\limits_{1}^{e}{x\text{dx}=\dfrac{{{e}^{2}}}{2}-\dfrac{{{x}^{2}}}{4}\mathop{|}_{0}^{e}}}=\dfrac{{{e}^{2}}}{2}-\dfrac{{{e}^{2}}}{4}+\dfrac{1}{4}=\dfrac{{{e}^{2}}+1}{4}.$
Đáp án D.