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Câu hỏi: Biết rằng tồn tại duy nhất bộ các số nguyên a, b, c sao cho $\int\limits_{2}^{3}{(4\text{x}+2)\ln \text{xdx}}=a+b\ln 2+c\ln 3$. Giá trị của $a+b+c$ bằng:
A. 19
B. $-19$
C. 5
D. $-5$
A. 19
B. $-19$
C. 5
D. $-5$
Đặt $I=\int\limits_{2}^{3}{(4\text{x}+2)\ln x\text{dx}}$
Đặt $\left\{ \begin{aligned}
& u=\ln \text{x} \\
& dv=(4\text{x}+2)d\text{x} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& du=\dfrac{d\text{x}}{x} \\
& v=2{{\text{x}}^{2}}+2\text{x}=2\text{x}(x+1) \\
\end{aligned} \right.$
$\Rightarrow I=\left. \left[ 2\text{x}(x+1)\ln \text{x} \right] \right|_{2}^{3}-\int\limits_{2}^{3}{\dfrac{2\text{x}(x+1)d\text{x}}{x}}=24\ln 3-12\ln 2-2\int\limits_{2}^{3}{(x+1)d\text{x}}$
$=24\ln 3-12\ln 2-\left. 2\left( \dfrac{{{x}^{2}}}{2}+x \right) \right|_{2}^{3}=24\ln 3-12\ln 2-2\left( \dfrac{15}{2}-4 \right)$
$=24\ln 3-12\ln 2-7=a+b\ln 2+c\ln 3$.
$\Rightarrow \left\{ \begin{aligned}
& a=-7 \\
& b=-12 \\
& c=24 \\
\end{aligned} \right.\Rightarrow a+b+c=-7-12+24=5$
Đặt $\left\{ \begin{aligned}
& u=\ln \text{x} \\
& dv=(4\text{x}+2)d\text{x} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& du=\dfrac{d\text{x}}{x} \\
& v=2{{\text{x}}^{2}}+2\text{x}=2\text{x}(x+1) \\
\end{aligned} \right.$
$\Rightarrow I=\left. \left[ 2\text{x}(x+1)\ln \text{x} \right] \right|_{2}^{3}-\int\limits_{2}^{3}{\dfrac{2\text{x}(x+1)d\text{x}}{x}}=24\ln 3-12\ln 2-2\int\limits_{2}^{3}{(x+1)d\text{x}}$
$=24\ln 3-12\ln 2-\left. 2\left( \dfrac{{{x}^{2}}}{2}+x \right) \right|_{2}^{3}=24\ln 3-12\ln 2-2\left( \dfrac{15}{2}-4 \right)$
$=24\ln 3-12\ln 2-7=a+b\ln 2+c\ln 3$.
$\Rightarrow \left\{ \begin{aligned}
& a=-7 \\
& b=-12 \\
& c=24 \\
\end{aligned} \right.\Rightarrow a+b+c=-7-12+24=5$
Đáp án C.